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Merge two BSTs with limited extra space

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Given two Binary Search Trees(BST), print the inorder traversal of merged BSTs. 

Examples:

Input: 
First BST 
       3
    /     \
 1       5
Second BST
     4
  /   \
2     6
Output: 1 2 3 4 5 6

Input:
First BST 
            8
         /  \
      2   10
    /
 1
Second BST 
          5
        / 
     3  
   /
0
Output: 0 1 2 3 5 8 10 

Recommended Practice

Merge two BSTs using Iterative Inorder Traversal:

The idea is to use iterative inorder traversal

Follow the steps below to solve the problem:

  • Consider two stacks s1 and s2 which stores the elements of the two trees.
  • Store the left view value of a tree1 in s1 and of tree2 in s2.
  • Compare the top values present in the stack and push the value accordingly in the result vector.
    • If s2 is empty then pop s1 and put the popped node value in the answer vector 
    • Else if both s1 and s2 are not empty then compare their top nodes’ value if s1.top()->val <= s2.top()->val then in this case push the s1.top()->val in the result vector and push its right child in the stack s1.
    • If s1 is empty then pop s2 and put the popped node value in the answer vector.
    • Else if both s1 and s2 are not empty then compare their top nodes’ value if s2.top()->val >= s1.top()->val then in this case push the s2.top()->val in the result vector and push its right child in the stack s2 
    • Loop while there are nodes not yet printed. The nodes may be in the stack(explored, but not printed) or maybe not yet explored 

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Structure of a BST Node
class node {
public:
    int data;
    node* left;
    node* right;
};
 
//.................... START OF STACK RELATED
//STUFF....................
// A stack node
class snode {
public:
    node* t;
    snode* next;
};
 
// Function to add an element k to stack
void push(snode** s, node* k)
{
    snode* tmp = new snode();
 
    // perform memory check here
    tmp->t = k;
    tmp->next = *s;
    (*s) = tmp;
}
 
// Function to pop an element t from stack
node* pop(snode** s)
{
    node* t;
    snode* st;
    st = *s;
    (*s) = (*s)->next;
    t = st->t;
    free(st);
    return t;
}
 
// Function to check whether the stack is empty or not
int isEmpty(snode* s)
{
    if (s == NULL)
        return 1;
 
    return 0;
}
//.................... END OF STACK RELATED
//STUFF....................
 
/* Utility function to create a new Binary Tree node */
node* newNode(int data)
{
    node* temp = new node;
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
 
/* A utility function to print Inorder traversal of a Binary
 * Tree */
void inorder(node* root)
{
    if (root != NULL) {
        inorder(root->left);
        cout << root->data << " ";
        inorder(root->right);
    }
}
 
// The function to print data of two BSTs in sorted order
void merge(node* root1, node* root2)
{
    // s1 is stack to hold nodes of first BST
    snode* s1 = NULL;
 
    // Current node of first BST
    node* current1 = root1;
 
    // s2 is stack to hold nodes of second BST
    snode* s2 = NULL;
 
    // Current node of second BST
    node* current2 = root2;
 
    // If first BST is empty, then output is inorder
    // traversal of second BST
    if (root1 == NULL) {
        inorder(root2);
        return;
    }
    // If second BST is empty, then output is inorder
    // traversal of first BST
    if (root2 == NULL) {
        inorder(root1);
        return;
    }
 
    // Run the loop while there are nodes not yet printed.
    // The nodes may be in stack(explored, but not printed)
    // or may be not yet explored
    while (current1 != NULL || !isEmpty(s1)
           || current2 != NULL || !isEmpty(s2)) {
        // Following steps follow iterative Inorder
        // Traversal
        if (current1 != NULL || current2 != NULL) {
            // Reach the leftmost node of both BSTs and push
            // ancestors of leftmost nodes to stack s1 and
            // s2 respectively
            if (current1 != NULL) {
                push(&s1, current1);
                current1 = current1->left;
            }
            if (current2 != NULL) {
                push(&s2, current2);
                current2 = current2->left;
            }
        }
        else {
            // If we reach a NULL node and either of the
            // stacks is empty, then one tree is exhausted,
            // print the other tree
            if (isEmpty(s1)) {
                while (!isEmpty(s2)) {
                    current2 = pop(&s2);
                    current2->left = NULL;
                    inorder(current2);
                }
                return;
            }
            if (isEmpty(s2)) {
                while (!isEmpty(s1)) {
                    current1 = pop(&s1);
                    current1->left = NULL;
                    inorder(current1);
                }
                return;
            }
 
            // Pop an element from both stacks and compare
            // the popped elements
            current1 = pop(&s1);
            current2 = pop(&s2);
 
            // If element of first tree is smaller, then
            // print it and push the right subtree. If the
            // element is larger, then we push it back to
            // the corresponding stack.
            if (current1->data < current2->data) {
                cout << current1->data << " ";
                current1 = current1->right;
                push(&s2, current2);
                current2 = NULL;
            }
            else {
                cout << current2->data << " ";
                current2 = current2->right;
                push(&s1, current1);
                current1 = NULL;
            }
        }
    }
}
 
/* Driver program to test above functions */
int main()
{
    node *root1 = NULL, *root2 = NULL;
 
    /* Let us create the following tree as first tree
            3
        / \
        1 5
    */
    root1 = newNode(3);
    root1->left = newNode(1);
    root1->right = newNode(5);
 
    /* Let us create the following tree as second tree
            4
        / \
        2 6
    */
    root2 = newNode(4);
    root2->left = newNode(2);
    root2->right = newNode(6);
 
    // Print sorted nodes of both trees
    merge(root1, root2);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

                    

C

#include <stdio.h>
#include <stdlib.h>
 
// Structure of a BST Node
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
//.................... START OF STACK RELATED
//STUFF....................
// A stack node
struct snode {
    struct node* t;
    struct snode* next;
};
 
// Function to add an element k to stack
void push(struct snode** s, struct node* k)
{
    struct snode* tmp
        = (struct snode*)malloc(sizeof(struct snode));
 
    // perform memory check here
    tmp->t = k;
    tmp->next = *s;
    (*s) = tmp;
}
 
// Function to pop an element t from stack
struct node* pop(struct snode** s)
{
    struct node* t;
    struct snode* st;
    st = *s;
    (*s) = (*s)->next;
    t = st->t;
    free(st);
    return t;
}
 
// Function to check whether the stack is empty or not
int isEmpty(struct snode* s)
{
    if (s == NULL)
        return 1;
 
    return 0;
}
//.................... END OF STACK RELATED
//STUFF....................
 
/* Utility function to create a new Binary Tree node */
struct node* newNode(int data)
{
    struct node* temp
        = (struct node*)malloc(sizeof(struct node));
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
 
/* A utility function to print Inorder traversal of a Binary
 * Tree */
void inorder(struct node* root)
{
    if (root != NULL) {
        inorder(root->left);
        printf("%d ", root->data);
        inorder(root->right);
    }
}
 
// The function to print data of two BSTs in sorted order
void merge(struct node* root1, struct node* root2)
{
    // s1 is stack to hold nodes of first BST
    struct snode* s1 = NULL;
 
    // Current node of first BST
    struct node* current1 = root1;
 
    // s2 is stack to hold nodes of second BST
    struct snode* s2 = NULL;
 
    // Current node of second BST
    struct node* current2 = root2;
 
    // If first BST is empty, then output is inorder
    // traversal of second BST
    if (root1 == NULL) {
        inorder(root2);
        return;
    }
    // If second BST is empty, then output is inorder
    // traversal of first BST
    if (root2 == NULL) {
        inorder(root1);
        return;
    }
 
    // Run the loop while there are nodes not yet printed.
    // The nodes may be in stack(explored, but not printed)
    // or may be not yet explored
    while (current1 != NULL || !isEmpty(s1)
           || current2 != NULL || !isEmpty(s2)) {
        // Following steps follow iterative Inorder
        // Traversal
        if (current1 != NULL || current2 != NULL) {
            // Reach the leftmost node of both BSTs and push
            // ancestors of leftmost nodes to stack s1 and
            // s2 respectively
            if (current1 != NULL) {
                push(&s1, current1);
                current1 = current1->left;
            }
            if (current2 != NULL) {
                push(&s2, current2);
                current2 = current2->left;
            }
        }
        else {
            // If we reach a NULL node and either of the
            // stacks is empty, then one tree is exhausted,
            // print the other tree
            if (isEmpty(s1)) {
                while (!isEmpty(s2)) {
                    current2 = pop(&s2);
                    current2->left = NULL;
                    inorder(current2);
                }
                return;
            }
            if (isEmpty(s2)) {
                while (!isEmpty(s1)) {
                    current1 = pop(&s1);
                    current1->left = NULL;
                    inorder(current1);
                }
                return;
            }
 
            // Pop an element from both stacks and compare
            // the popped elements
            current1 = pop(&s1);
            current2 = pop(&s2);
 
            // If element of first tree is smaller, then
            // print it and push the right subtree. If the
            // element is larger, then we push it back to
            // the corresponding stack.
            if (current1->data < current2->data) {
                printf("%d ", current1->data);
                current1 = current1->right;
                push(&s2, current2);
                current2 = NULL;
            }
            else {
                printf("%d ", current2->data);
                current2 = current2->right;
                push(&s1, current1);
                current1 = NULL;
            }
        }
    }
}
 
/* Driver program to test above functions */
int main()
{
    struct node *root1 = NULL, *root2 = NULL;
 
    /* Let us create the following tree as first tree
            3
          /  \
         1    5
     */
    root1 = newNode(3);
    root1->left = newNode(1);
    root1->right = newNode(5);
 
    /* Let us create the following tree as second tree
            4
          /  \
         2    6
     */
    root2 = newNode(4);
    root2->left = newNode(2);
    root2->right = newNode(6);
 
    // Print sorted nodes of both trees
    merge(root1, root2);
 
    return 0;
}

                    

Java

public class Merge2BST {
 
    /* A utility function to print
    Inorder traversal of a Binary Tree */
    static void inorder(Node root)
    {
        if (root != null) {
            inorder(root.left);
            System.out.print(root.data + " ");
            inorder(root.right);
        }
    }
 
    // The function to print data of two BSTs in sorted
    // order
    static void merge(Node root1, Node root2)
    {
        // s1 is stack to hold nodes of first BST
        SNode s1 = new SNode();
 
        // Current node of first BST
        Node current1 = root1;
 
        // s2 is stack to hold nodes of second BST
        SNode s2 = new SNode();
 
        // Current node of second BST
        Node current2 = root2;
 
        // If first BST is empty, then output is inorder
        // traversal of second BST
        if (root1 == null) {
            inorder(root2);
            return;
        }
 
        // If second BST is empty, then output is inorder
        // traversal of first BST
        if (root2 == null) {
            inorder(root1);
            return;
        }
 
        // Run the loop while there are nodes not yet
        // printed. The nodes may be in stack(explored, but
        // not printed) or may be not yet explored
        while (current1 != null || !s1.isEmpty()
               || current2 != null || !s2.isEmpty()) {
 
            // Following steps follow iterative Inorder
            // Traversal
            if (current1 != null || current2 != null) {
                // Reach the leftmost node of both BSTs and
                // push ancestors of leftmost nodes to stack
                // s1 and s2 respectively
                if (current1 != null) {
 
                    s1.push(current1);
                    current1 = current1.left;
                }
                if (current2 != null) {
                    s2.push(current2);
                    current2 = current2.left;
                }
            }
            else {
 
                // If we reach a NULL node and either of the
                // stacks is empty, then one tree is
                // exhausted, print the other tree
                if (s1.isEmpty()) {
                    while (!s2.isEmpty()) {
                        current2 = s2.pop();
                        current2.left = null;
                        inorder(current2);
                    }
                    return;
                }
                if (s2.isEmpty()) {
                    while (!s1.isEmpty()) {
                        current1 = s1.pop();
                        current1.left = null;
                        inorder(current1);
                    }
                    return;
                }
 
                // Pop an element from both stacks and
                // compare the popped elements
                current1 = s1.pop();
 
                current2 = s2.pop();
 
                // If element of first tree is smaller, then
                // print it and push the right subtree. If
                // the element is larger, then we push it
                // back to the corresponding stack.
                if (current1.data < current2.data) {
                    System.out.print(current1.data + " ");
                    current1 = current1.right;
                    s2.push(current2);
                    current2 = null;
                }
                else {
                    System.out.print(current2.data + " ");
                    current2 = current2.right;
                    s1.push(current1);
                    current1 = null;
                }
            }
        }
        System.out.println(s1.t);
        System.out.println(s2.t);
    }
 
    /* Driver code */
    public static void main(String[] args)
    {
        Node root1 = null, root2 = null;
 
        /* Let us create the following tree as first tree
                3
            / \
            1 5
        */
        root1 = new Node(3);
        root1.left = new Node(1);
        root1.right = new Node(5);
 
        /* Let us create the following tree as second tree
                4
            / \
            2 6
        */
        root2 = new Node(4);
        root2.left = new Node(2);
        root2.right = new Node(6);
 
        // Print sorted nodes of both trees
        merge(root1, root2);
    }
}
 
// Structure of a BST Node
class Node {
 
    int data;
    Node left;
    Node right;
    public Node(int data)
    {
        // TODO Auto-generated constructor stub
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// A stack node
class SNode {
    SNode head;
    Node t;
    SNode next;
 
    // Function to add an element k to stack
    void push(Node k)
    {
        SNode tmp = new SNode();
 
        // Perform memory check here
        tmp.t = k;
        tmp.next = this.head;
        this.head = tmp;
    }
 
    // Function to pop an element t from stack
    Node pop()
    {
 
        SNode st;
        st = this.head;
        head = head.next;
 
        return st.t;
    }
 
    // Function to check whether the stack is empty or not
    boolean isEmpty()
    {
        if (this.head == null)
            return true;
 
        return false;
    }
}
 
// This code is contributed by nidhisebastian008

                    

Python 3

# Class to create a new Tree Node
class newNode:
    def __init__(self, data: int):
        self.data = data
        self.left = None
        self.right = None
 
 
def inorder(root: newNode):
 
    if root:
        inorder(root.left)
        print(root.data, end=" ")
        inorder(root.right)
 
 
def merge(root1: newNode, root2: newNode):
 
    # s1 is stack to hold nodes of first BST
    s1 = []
 
    # Current node of first BST
    current1 = root1
 
    # s2 is stack to hold nodes of first BST
    s2 = []
 
    # Current node of second BST
    current2 = root2
 
    # If first BST is empty then the output is the
    # inorder traversal of the second BST
    if not root1:
        return inorder(root2)
 
    # If the second BST is empty then the output is the
    # inorder traversal of the first BST
    if not root2:
        return inorder(root1)
 
    # Run the loop while there are nodes not yet printed.
    # The nodes may be in stack(explored, but not printed)
    # or may be not yet explored
    while current1 or s1 or current2 or s2:
 
        # Following steps follow iterative Inorder Traversal
        if current1 or current2:
 
            # Reach the leftmost node of both BSTs and push ancestors of
            # leftmost nodes to stack s1 and s2 respectively
            if current1:
                s1.append(current1)
                current1 = current1.left
 
            if current2:
                s2.append(current2)
                current2 = current2.left
 
        else:
 
            # If we reach a NULL node and either of the stacks is empty,
            # then one tree is exhausted, print the other tree
 
            if not s1:
                while s2:
                    current2 = s2.pop()
                    current2.left = None
                    inorder(current2)
                    return
            if not s2:
                while s1:
                    current1 = s1.pop()
                    current1.left = None
                    inorder(current1)
                    return
 
            # Pop an element from both stacks and compare the
            # popped elements
            current1 = s1.pop()
            current2 = s2.pop()
 
            # If element of first tree is smaller, then print it
            # and push the right subtree. If the element is larger,
            # then we push it back to the corresponding stack.
            if current1.data < current2.data:
                print(current1.data, end=" ")
                current1 = current1.right
                s2.append(current2)
                current2 = None
 
            else:
                print(current2.data, end=" ")
                current2 = current2.right
                s1.append(current1)
                current1 = None
 
# Driver code
 
 
def main():
 
    # Let us create the following tree as first tree
    #     3
    #     / \
    # 1 5
 
    root1 = newNode(3)
    root1.left = newNode(1)
    root1.right = newNode(5)
 
    # Let us create the following tree as second tree
    #     4
    #     / \
    # 2 6
    #
 
    root2 = newNode(4)
    root2.left = newNode(2)
    root2.right = newNode(6)
 
    merge(root1, root2)
 
 
if __name__ == "__main__":
    main()
 
# This code is contributed by Koushik Reddy Bukkasamudram

                    

C#

// C# program to implement the
// above approach
using System;
class Merge2BST {
 
    /* A utility function to print
       Inorder traversal of a Binary
       Tree */
    static void inorder(Node root)
    {
        if (root != null) {
            inorder(root.left);
            Console.Write(root.data + " ");
            inorder(root.right);
        }
    }
 
    // The function to print data
    // of two BSTs in sorted order
    static void merge(Node root1, Node root2)
    {
        // s1 is stack to hold nodes
        // of first BST
        SNode s1 = new SNode();
 
        // Current node of first BST
        Node current1 = root1;
 
        // s2 is stack to hold nodes
        // of second BST
        SNode s2 = new SNode();
 
        // Current node of second BST
        Node current2 = root2;
 
        // If first BST is empty, then
        // output is inorder traversal
        // of second BST
        if (root1 == null) {
            inorder(root2);
            return;
        }
 
        // If second BST is empty,
        // then output is inorder
        // traversal of first BST
        if (root2 == null) {
            inorder(root1);
            return;
        }
 
        // Run the loop while there
        // are nodes not yet printed.
        // The nodes may be in stack
        // (explored, but not printed)
        // or may be not yet explored
        while (current1 != null || !s1.isEmpty()
               || current2 != null || !s2.isEmpty()) {
            // Following steps follow
            // iterative Inorder Traversal
            if (current1 != null || current2 != null) {
                // Reach the leftmost node of
                // both BSTs and push ancestors
                // of leftmost nodes to stack
                // s1 and s2 respectively
                if (current1 != null) {
                    s1.push(current1);
                    current1 = current1.left;
                }
                if (current2 != null) {
                    s2.push(current2);
                    current2 = current2.left;
                }
            }
            else {
                // If we reach a NULL node and
                // either of the stacks is empty,
                // then one tree is exhausted,
                // print the other tree
                if (s1.isEmpty()) {
                    while (!s2.isEmpty()) {
                        current2 = s2.pop();
                        current2.left = null;
                        inorder(current2);
                    }
                    return;
                }
                if (s2.isEmpty()) {
                    while (!s1.isEmpty()) {
                        current1 = s1.pop();
                        current1.left = null;
                        inorder(current1);
                    }
                    return;
                }
 
                // Pop an element from both
                // stacks and compare the
                // popped elements
                current1 = s1.pop();
 
                current2 = s2.pop();
 
                // If element of first tree is
                // smaller, then print it
                // and push the right subtree.
                // If the element is larger,
                // then we push it back to the
                // corresponding stack.
                if (current1.data < current2.data) {
                    Console.Write(current1.data + " ");
                    current1 = current1.right;
                    s2.push(current2);
                    current2 = null;
                }
                else {
                    Console.Write(current2.data + " ");
                    current2 = current2.right;
                    s1.push(current1);
                    current1 = null;
                }
            }
        }
        Console.Write(s1.t + "\n");
        Console.Write(s2.t + "\n");
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        Node root1 = null, root2 = null;
 
        /* Let us create the
           following tree as
           first tree
 
                   3
                  / \
                  1 5
        */
        root1 = new Node(3);
        root1.left = new Node(1);
        root1.right = new Node(5);
 
        /* Let us create the following
           tree as second tree
 
                   4
                  / \
                  2 6
        */
        root2 = new Node(4);
        root2.left = new Node(2);
        root2.right = new Node(6);
 
        // Print sorted nodes of
        // both trees
        merge(root1, root2);
    }
}
 
// Structure of a BST Node
class Node {
 
    public int data;
    public Node left;
    public Node right;
 
    public Node(int data)
    {
        // TODO Auto-generated
        // constructor stub
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// A stack node
class SNode {
 
    SNode head;
    public Node t;
    SNode next;
 
    // Function to add an element
    // k to stack
    public void push(Node k)
    {
        SNode tmp = new SNode();
 
        // Perform memory check here
        tmp.t = k;
        tmp.next = this.head;
        this.head = tmp;
    }
 
    // Function to pop an element
    // t from stack
    public Node pop()
    {
        SNode st;
        st = this.head;
        head = head.next;
 
        return st.t;
    }
 
    // Function to check whether
    // the stack is empty or not
    public bool isEmpty()
    {
        if (this.head == null)
            return true;
 
        return false;
    }
}
 
// This code is contributed by Rutvik_56

                    

Javascript

<script>
 
      // JavaScript program to implement the
      // above approach
 
      // Structure of a BST Node
      class Node {
        constructor(data) {
          // TODO Auto-generated
          // constructor stub
          this.data = data;
          this.left = null;
          this.right = null;
        }
      }
 
      // A stack node
      class SNode {
        constructor() {
          this.head = null;
          this.t = null;
          this.next = null;
        }
 
        // Function to add an element
        // k to stack
        push(k) {
          var tmp = new SNode();
 
          // Perform memory check here
          tmp.t = k;
          tmp.next = this.head;
          this.head = tmp;
        }
 
        // Function to pop an element
        // t from stack
        pop() {
          var st;
          st = this.head;
          this.head = this.head.next;
 
          return st.t;
        }
 
        // Function to check whether
        // the stack is empty or not
        isEmpty() {
          if (this.head == null) return true;
 
          return false;
        }
      }
 
      /* A utility function to print
        Inorder traversal of a Binary
        Tree */
      function inorder(root) {
        if (root != null) {
          inorder(root.left);
          document.write(root.data + " ");
          inorder(root.right);
        }
      }
 
      // The function to print data
      // of two BSTs in sorted order
      function merge(root1, root2) {
        // s1 is stack to hold nodes
        // of first BST
        var s1 = new SNode();
 
        // Current node of first BST
        var current1 = root1;
 
        // s2 is stack to hold nodes
        // of second BST
        var s2 = new SNode();
 
        // Current node of second BST
        var current2 = root2;
 
        // If first BST is empty, then
        // output is inorder traversal
        // of second BST
        if (root1 == null) {
          inorder(root2);
          return;
        }
 
        // If second BST is empty,
        // then output is inorder
        // traversal of first BST
        if (root2 == null) {
          inorder(root1);
          return;
        }
 
        // Run the loop while there
        // are nodes not yet printed.
        // The nodes may be in stack
        // (explored, but not printed)
        // or may be not yet explored
        while (
          current1 != null ||
          !s1.isEmpty() ||
          current2 != null ||
          !s2.isEmpty()
        ) {
          // Following steps follow
          // iterative Inorder Traversal
          if (current1 != null || current2 != null) {
            // Reach the leftmost node of
            // both BSTs and push ancestors
            // of leftmost nodes to stack
            // s1 and s2 respectively
            if (current1 != null) {
              s1.push(current1);
              current1 = current1.left;
            }
            if (current2 != null) {
              s2.push(current2);
              current2 = current2.left;
            }
          } else {
            // If we reach a NULL node and
            // either of the stacks is empty,
            // then one tree is exhausted,
            // print the other tree
            if (s1.isEmpty()) {
              while (!s2.isEmpty()) {
                current2 = s2.pop();
                current2.left = null;
                inorder(current2);
              }
              return;
            }
            if (s2.isEmpty()) {
              while (!s1.isEmpty()) {
                current1 = s1.pop();
                current1.left = null;
                inorder(current1);
              }
              return;
            }
 
            // Pop an element from both
            // stacks and compare the
            // popped elements
            current1 = s1.pop();
 
            current2 = s2.pop();
 
            // If element of first tree is
            // smaller, then print it
            // and push the right subtree.
            // If the element is larger,
            // then we push it back to the
            // corresponding stack.
            if (current1.data < current2.data) {
              document.write(current1.data + " ");
              current1 = current1.right;
              s2.push(current2);
              current2 = null;
            } else {
              document.write(current2.data + " ");
              current2 = current2.right;
              s1.push(current1);
              current1 = null;
            }
          }
        }
        document.write(s1.t + "<br>");
        document.write(s2.t + "<br>");
      }
 
      // Driver code
      var root1 = null,
        root2 = null;
 
      /* Let us create the
    following tree as
    first tree
     
            3
            / \
            1 5
*/
      root1 = new Node(3);
      root1.left = new Node(1);
      root1.right = new Node(5);
 
      /* Let us create the following
    tree as second tree
                 
            4
            / \
            2 6
*/
      root2 = new Node(4);
      root2.left = new Node(2);
      root2.right = new Node(6);
 
      // Print sorted nodes of
      // both trees
      merge(root1, root2);
       
</script>

                    

Output
1 2 3 4 5 6 

Time Complexity: O(M+N), M is the size of the first tree and N is the size of the second tree
Auxiliary Space: O(H1 + H2), H1 is the height of the first tree and H2 is the height of the second tree

Merge two BSTs using Inbuilt Stack Data structure:

In this method, we use the inbuilt stack that is present in the STL library so as to get rid of the implementation of the stack part of the code that has been done in the previous implementation.

Below is the implementation of the above approach.

C++

#include <bits/stdc++.h>
using namespace std;
 
// Structure of a BST Node
class Node {
public:
    int val;
    Node* left;
    Node* right;
};
 
/* Utility function to create a new Binary Tree Node */
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->val = data;
    temp->left = nullptr;
    temp->right = nullptr;
    return temp;
}
 
vector<int> mergeTwoBST(Node* root1, Node* root2)
{
    vector<int> res;
    stack<Node*> s1, s2;
    while (root1 || root2 || !s1.empty() || !s2.empty()) {
        while (root1) {
            s1.push(root1);
            root1 = root1->left;
        }
        while (root2) {
            s2.push(root2);
            root2 = root2->left;
        }
        // Step 3 Case 1:-
        if (s2.empty()
            || (!s1.empty()
                && s1.top()->val <= s2.top()->val)) {
            root1 = s1.top();
            s1.pop();
            res.push_back(root1->val);
            root1 = root1->right;
        }
        // Step 3 case 2 :-
        else {
            root2 = s2.top();
            s2.pop();
            res.push_back(root2->val);
            root2 = root2->right;
        }
    }
    return res;
}
 
/* Driver program to test above functions */
int main()
{
    Node *root1 = nullptr, *root2 = nullptr;
 
    /* Let us create the following tree as first tree
       3
      / \
      1 5
    */
    root1 = newNode(3);
    root1->left = newNode(1);
    root1->right = newNode(5);
 
    /* Let us create the following tree as second tree
       4
      / \
      2 6
    */
    root2 = newNode(4);
    root2->left = newNode(2);
    root2->right = newNode(6);
 
    // Print sorted Nodes of both trees
    vector<int> ans = mergeTwoBST(root1, root2);
    for (auto it : ans)
        cout << it << " ";
    return 0;
}
 
// This code is contributed by Aditya kumar (adityakumar129)

                    

Java

// Java Code for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    static void mergeTwoBST(Node root1, Node root2)
    {
        List<Integer> res = new ArrayList<Integer>();
        Stack<Node> s1 = new Stack<Node>();
        Stack<Node> s2 = new Stack<Node>();
 
        while (root1 != null || root2 != null
               || !s1.isEmpty() || !s2.isEmpty()) {
            while (root1 != null) {
                s1.push(root1);
                root1 = root1.left;
            }
            while (root2 != null) {
                s2.push(root2);
                root2 = root2.left;
            }
 
            if (s2.isEmpty()
                || (!s1.isEmpty()
                    && s1.peek().data <= s2.peek().data)) {
                root1 = s1.peek();
                s1.pop();
                res.add(root1.data);
                root1 = root1.right;
            }
            else {
                root2 = s2.peek();
                s2.pop();
                res.add(root2.data);
                root2 = root2.right;
            }
        }
        for (int i = 0; i < res.size(); i++) {
            System.out.print(res.get(i) + " ");
        }
    }
 
    public static void main(String[] args)
    {
        Node root1 = null, root2 = null;
 
        /* Let us create the following tree as first tree
                    3
                / \
                1 5
            */
        root1 = new Node(3);
        root1.left = new Node(1);
        root1.right = new Node(5);
 
        /* Let us create the following tree as second tree
                    4
                / \
                2 6
            */
        root2 = new Node(4);
        root2.left = new Node(2);
        root2.right = new Node(6);
 
        mergeTwoBST(root1, root2);
    }
}
 
// Structure of a BST Node
class Node {
 
    int data;
    Node left;
    Node right;
    public Node(int data)
    {
        // TODO Auto-generated constructor stub
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// This code is contributed by lokesh(lokeshmvs21).

                    

Python3

# Python program to Merge two BSTs with limited extra space
 
# Structure of a BST Node
 
 
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
 
 
def mergeTwoBST(root1, root2):
    res = []
    s1, s2 = [], []
 
    while root1 or root2 or s1 or s2:
        while root1:
            s1.append(root1)
            root1 = root1.left
 
        while root2:
            s2.append(root2)
            root2 = root2.left
 
        # Step 3 Case 1:-
        if not s2 or (s1 and s1[-1].val <= s2[-1].val):
            root1 = s1[-1]
            del s1[-1]
            res.append(root1.val)
            root1 = root1.right
 
        #  Step 3 case 2 :-
        else:
            root2 = s2[-1]
            del s2[-1]
            res.append(root2.val)
            root2 = root2.right
 
    return res
 
 
# Driver program to test above functions
if __name__ == '__main__':
    root1 = None
    root2 = None
 
    '''
    Let us create the following tree as first tree
             3
            / \
           1   5
    '''
 
    root1 = Node(3)
    root1.left = Node(1)
    root1.right = Node(5)
 
    '''
    Let us create the following tree as second tree
             4
            / \
           2   6
    '''
 
    root2 = Node(4)
    root2.left = Node(2)
    root2.right = Node(6)
 
    ans = mergeTwoBST(root1, root2)
    for x in ans:
        print(x, end=" ")
 
# This code is contributed by Tapesh(tapeshdua420)

                    

C#

// C# Code for the above approach
using System;
using System.Collections.Generic;
 
// Structure of a BST Node
class Node {
 
    public int data;
    public Node left;
    public Node right;
 
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
class GFG {
 
    static void mergeTwoBST(Node root1, Node root2)
    {
 
        List<int> res = new List<int>();
        Stack<Node> s1 = new Stack<Node>();
        Stack<Node> s2 = new Stack<Node>();
 
        while (root1 != null || root2 != null
               || s1.Count != 0 || s2.Count != 0) {
            while (root1 != null) {
                s1.Push(root1);
                root1 = root1.left;
            }
 
            while (root2 != null) {
                s2.Push(root2);
                root2 = root2.left;
            }
 
            if (s2.Count == 0
                || (s1.Count != 0
                    && s1.Peek().data <= s2.Peek().data)) {
                root1 = s1.Peek();
                s1.Pop();
                res.Add(root1.data);
                root1 = root1.right;
            }
            else {
                root2 = s2.Peek();
                s2.Pop();
                res.Add(root2.data);
                root2 = root2.right;
            }
        }
        for (int i = 0; i < res.Count; i++) {
            Console.Write(res[i] + " ");
        }
    }
    public static void Main(String[] args)
    {
        Node root1 = null, root2 = null;
 
        /* Let us create the following tree as first tree
              3
            / \
           1  5
        */
        root1 = new Node(3);
        root1.left = new Node(1);
        root1.right = new Node(5);
 
        /* Let us create the following tree as second tree
              4
            / \
           2  6
        */
        root2 = new Node(4);
        root2.left = new Node(2);
        root2.right = new Node(6);
 
        mergeTwoBST(root1, root2);
    }
}
// This code is contributed by Tapesh (tapeshdua420)

                    

Javascript

// JavaScript Code for the above approach
     
    function Node(data, left = null, right = null) {
        this.data = data;
        this.left = left;
        this.right = right;
    }
 
    function mergeTwoBST(root1, root2) {
        let res = [];
        let s1 = [];
          let s2 = [];
 
        while (root1 !== null || root2 !== null || s1.length !== 0 || s2.length !== 0) {
            while (root1 !== null) {
                s1.push(root1);
                root1 = root1.left;
            }
            while (root2 !== null) {
                s2.push(root2);
                root2 = root2.left;
            }
 
            if (s2.length === 0 || (s1.length !== 0 && s1[s1.length - 1].data <= s2[s2.length - 1].data)) {
                root1 = s1.pop();
                res.push(root1.data);
                root1 = root1.right;
            }
            else {
                root2 = s2.pop();
                res.push(root2.data);
                root2 = root2.right;
            }
        }
 
          console.log(res.join(' '));
    }
 
    function main() {
        let root1 = null;
        let root2 = null;
 
          /* Create the following tree as first tree
                    3
                   / \
                  1   5
            */
          root1 = new Node(3, new Node(1), new Node(5));
 
          /* Create the following tree as second tree
                    4
                      / \
                    2   6
            */
          root2 = new Node(4, new Node(2), new Node(6));
 
          mergeTwoBST(root1, root2);
    }
 
    main();
     
    // This code is contributed by lokesh.

                    

Output
1 2 3 4 5 6 

Time Complexity: O(M+N), M is the size of the first tree and N is the size of the second tree
Auxiliary Space: O(H1 + H2), H1 is the height of the first tree and H2 is the height of the second tree

Merge two BSTs using Doubly Linked List 

Follow the steps below to solve the problem:

Time Complexity: O(N),N is the size of the tree
Auxiliary Space: O(H),H is the height of the tree 

Time Complexity: O(M+N), M is the size of the first tree and N is the size of the second tree
Auxiliary Space: O(1)

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
class Node {
public:
    int val;
    Node* left;
    Node* right;
};
 
/* Utility function to create a new Binary Tree Node */
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->val = data;
    temp->left = nullptr;
    temp->right = nullptr;
    return temp;
}
 
/* UTILITY FUNCTIONS */
/* MoveNode() function takes the
node from the front of the source,
and move it to the front of the dest.
It is an error to call this with the
source list empty.
  
Before calling MoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
  
After calling MoveNode():
source == {2, 3}
dest == {1, 1, 2, 3} */
void MoveNode(Node** destRef, Node** sourceRef)
{
    /* the front source node */
    Node* newNode = *sourceRef;
    assert(newNode != NULL);
  
    /* Advance the source pointer */
    *sourceRef = newNode->right;
  
    /* Link the old dest of the new node */
    newNode->right = *destRef;
  
    /* Move dest to point to the new node */
    *destRef = newNode;
}
 
 
/* Merge sorted Double linked list */
Node* mergeLinkedList(Node* a, Node* b)
{
      Node dummy;
  
    /* tail points to the last result node */
    Node* tail = &dummy;
  
    /* so tail->next is the place to
    add new nodes to the result. */
    dummy.right = NULL;
    while (1) {
        if (a == NULL) {
            /* if either list runs out, use the
            other list */
            tail->right = b;
            break;
        }
        else if (b == NULL) {
            tail->right = a;
            break;
        }
        if (a->val <= b->val)
            MoveNode(&(tail->right), &a);
        else
            MoveNode(&(tail->right), &b);
  
        tail = tail->right;
    }
    return (dummy.right);
}
 
/* convert sorted Double linked list */
void convertIntoSortedDLL(Node* root, Node*& head)
{
      //base case
    if (root == NULL)
        return;
   
    convertIntoSortedDLL(root->right, head);
    root->right = head;
   
    if (head != NULL)
        head->left = root;
   
    head = root;
   
    convertIntoSortedDLL(root->left, head);
}
 
/* Function to print nodes in a given linked list */
void printList(Node* head)
{
    while (head) {
        printf("%d ", head->val);
        head = head->right;
    }
}
 
int main()
{
    Node *root1 = nullptr, *root2 = nullptr;
 
    /* Let us create the following tree as first tree
       3
      / \
      1 5
    */
    root1 = newNode(3);
    root1->left = newNode(1);
    root1->right = newNode(5);
 
    /* Let us create the following tree as second tree
       4
      / \
      2 6
    */
    root2 = newNode(4);
    root2->left = newNode(2);
    root2->right = newNode(6);
 
    // Convert BST into sorted DLL
      Node*head1=NULL;
    Node*head2=NULL;
    convertIntoSortedDLL(root1,head1);
    convertIntoSortedDLL(root2,head2);
 
    // merge sorted DLL
    Node* ans = mergeLinkedList(head1, head2);
     
    printList(ans);
    return 0;
  // This code is contributed by Ajay Rawat(theajayrawat)
}

                    

Java

// Java code addition
 
import java.io.*;
import java.util.*;
 
class Node {
    int data;
    Node left, right;
 
    /* Utility function to create a new Binary Tree Node */
    public Node(int data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
class BinaryTree {
    Node root;
 
    public void insert(int data) {
        root = insertRec(root, data);
    }
 
    public Node insertRec(Node root, int data) {
        if (root == null) {
            root = new Node(data);
            return root;
        }
 
        if (data < root.data) {
            root.left = insertRec(root.left, data);
        } else if (data > root.data) {
            root.right = insertRec(root.right, data);
        }
 
        return root;
    }
 
    /* convert sorted Double linked list */
    public void convertIntoSortedDLL(Node root, Node[] head) {
        if (root == null) {
            return;
        }
 
        convertIntoSortedDLL(root.right, head);
        root.right = head[0];
 
        if (head[0] != null) {
            head[0].left = root;
        }
 
        head[0] = root;
 
        convertIntoSortedDLL(root.left, head);
    }
 
    /* Merge sorted Double linked list */
    public Node mergeLinkedList(Node head1, Node head2) {
        Node dummy = new Node(0);
         
        /* tail points to the last result node */
        Node tail = dummy;
 
        /* so tail->next is the place to
      add new nodes to the result. */
        while (true) {
            if (head1 == null) {
                 
                /* if either list runs out, use the
              other list */
                tail.right = head2;
                break;
            }
 
            if (head2 == null) {
                tail.right = head1;
                break;
            }
 
            if (head1.data <= head2.data) {
                tail.right = head1;
                head1.left = tail;
                head1 = head1.right;
            } else {
                tail.right = head2;
                head2.left = tail;
                head2 = head2.right;
            }
 
            tail = tail.right;
        }
 
        Node res = dummy.right;
        res.left = null;
        return res;
    }
 
    /* Function to print nodes in a given linked list */
    public void printList(Node node) {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.right;
        }
    }
}
 
public class Main {
    public static void main(String[] args) {
        BinaryTree tree1 = new BinaryTree();
        BinaryTree tree2 = new BinaryTree();
 
        /* Let us create the following tree as first tree
               3
              / \
             1   5
        */
        tree1.insert(3);
        tree1.insert(1);
        tree1.insert(5);
 
        /* Let us create the following tree as second tree
               4
              / \
             2   6
        */
        tree2.insert(4);
        tree2.insert(2);
        tree2.insert(6);
 
        // Convert BST into sorted DLL
        Node[] head1 = {null};
        Node[] head2 = {null};
 
        tree1.convertIntoSortedDLL(tree1.root, head1);
        tree2.convertIntoSortedDLL(tree2.root, head2);
 
        // merge sorted DLL
        Node ans = tree1.mergeLinkedList(head1[0], head2[0]);
 
        tree1.printList(ans);
    }
}
 
// The code is contributed by Arushi Goel.

                    

Python3

class Node:
    def __init__(self, data):
      # Utility function to create a new Binary Tree Node
        self.data = data
        self.left = None
        self.right = None
 
 
class BinaryTree:
    def __init__(self):
        self.root = None
 
    def insert(self, data):
        self.root = self.insertRec(self.root, data)
 
    def insertRec(self, root, data):
        if root is None:
            root = Node(data)
            return root
 
        if data < root.data:
            root.left = self.insertRec(root.left, data)
        elif data > root.data:
            root.right = self.insertRec(root.right, data)
 
        return root
 
    # convert sorted Double linked list
    def convertIntoSortedDLL(self, root, head):
      # base case
        if root is None:
            return
 
        self.convertIntoSortedDLL(root.right, head)
        root.right = head[0]
 
        if head[0] is not None:
            head[0].left = root
 
        head[0] = root
 
        self.convertIntoSortedDLL(root.left, head)
 
    # Merge sorted Double linked list
    def mergeLinkedList(self, head1, head2):
        dummy = Node(0)
 
        # tail points to the last result node
        tail = dummy
 
        # so tail->next is the place to
      # add new nodes to the result.
        while True:
            if head1 is None:
              # if either list runs out, use the
              # other list
                tail.right = head2
                break
 
            if head2 is None:
                tail.right = head1
                break
 
            if head1.data <= head2.data:
                tail.right = head1
                head1.left = tail
                head1 = head1.right
            else:
                tail.right = head2
                head2.left = tail
                head2 = head2.right
 
            tail = tail.right
 
        res = dummy.right
        res.left = None
        return res
 
    # Function to print nodes in a given linked list
    def printList(self, node):
        while node is not None:
            print(node.data, end=" ")
            node = node.right
 
 
if __name__ == "__main__":
    tree1 = BinaryTree()
    tree2 = BinaryTree()
 
    # Let us create the following tree as first tree
    #       3
    #       / \
    #     1   5
 
    tree1.insert(3)
    tree1.insert(1)
    tree1.insert(5)
 
    # Let us create the following tree as second tree
    #         4
    #         / \
    #       2   6
 
    tree2.insert(4)
    tree2.insert(2)
    tree2.insert(6)
 
    # Convert BST into sorted DLL
    head1 = [None]
    head2 = [None]
 
    tree1.convertIntoSortedDLL(tree1.root, head1)
    tree2.convertIntoSortedDLL(tree2.root, head2)
 
    # merge sorted DLL
    ans = tree1.mergeLinkedList(head1[0], head2[0])
 
    tree1.printList(ans)

                    

C#

using System;
 
public class Node {
    public int data;
    public Node left;
    public Node right;
 
    /* Utility function to create a new Binary Tree Node */
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
public class BinaryTree {
    public Node root;
 
    public BinaryTree() { this.root = null; }
 
    public void insert(int data)
    {
        this.root = insertRec(this.root, data);
    }
 
    private Node insertRec(Node root, int data)
    {
        if (root == null) {
            root = new Node(data);
            return root;
        }
 
        if (data < root.data) {
            root.left = insertRec(root.left, data);
        }
        else if (data > root.data) {
            root.right = insertRec(root.right, data);
        }
 
        return root;
    }
 
    /* convert sorted Double linked list */
    public void convertIntoSortedDLL(Node root,
                                     ref Node head)
    {
        // base case
        if (root == null) {
            return;
        }
 
        convertIntoSortedDLL(root.right, ref head);
        root.right = head;
 
        if (head != null) {
            head.left = root;
        }
 
        head = root;
 
        convertIntoSortedDLL(root.left, ref head);
    }
 
    /* Merge sorted Double linked list */
    public Node mergeLinkedList(Node head1, Node head2)
    {
        Node dummy = new Node(0);
 
        /* tail points to the last result node */
        Node tail = dummy;
 
        /* so tail->next is the place to
      add new nodes to the result. */
        while (true) {
            if (head1 == null) {
                /* if either list runs out, use the
              other list */
                tail.right = head2;
                break;
            }
 
            if (head2 == null) {
                tail.right = head1;
                break;
            }
 
            if (head1.data <= head2.data) {
                tail.right = head1;
                head1.left = tail;
                head1 = head1.right;
            }
            else {
                tail.right = head2;
                head2.left = tail;
                head2 = head2.right;
            }
 
            tail = tail.right;
        }
 
        Node res = dummy.right;
        res.left = null;
        return res;
    }
 
    /* Function to print nodes in a given linked list */
    public void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.right;
        }
    }
}
 
public class Program {
    public static void Main()
    {
        BinaryTree tree1 = new BinaryTree();
        BinaryTree tree2 = new BinaryTree();
 
        /* Let us create the following tree as first tree
               3
              / \
             1   5
        */
        tree1.insert(3);
        tree1.insert(1);
        tree1.insert(5);
 
        /* Let us create the following tree as second tree
               4
              / \
             2   6
        */
        tree2.insert(4);
        tree2.insert(2);
        tree2.insert(6);
 
        // Convert BST into sorted DLL
        Node head1 = null;
        Node head2 = null;
 
        tree1.convertIntoSortedDLL(tree1.root, ref head1);
        tree2.convertIntoSortedDLL(tree2.root, ref head2);
 
        // merge sorted DLL
        Node ans = tree1.mergeLinkedList(head1, head2);
 
        tree1.printList(ans);
    }
}

                    

Javascript

// javascript code addition
 
class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}
 
class BinaryTree {
  constructor() {
    this.root = null;
  }
 
  insert(data) {
    this.root = this.insertRec(this.root, data);
  }
 
  insertRec(root, data) {
    if (root === null) {
      root = new Node(data);
      return root;
    }
 
    if (data < root.data) {
      root.left = this.insertRec(root.left, data);
    } else if (data > root.data) {
      root.right = this.insertRec(root.right, data);
    }
 
    return root;
  }
 
  convertIntoSortedDLL(root, head) {
    if (root === null) {
      return;
    }
 
    this.convertIntoSortedDLL(root.right, head);
    root.right = head[0];
 
    if (head[0] !== null) {
      head[0].left = root;
    }
 
    head[0] = root;
 
    this.convertIntoSortedDLL(root.left, head);
  }
 
  mergeLinkedList(head1, head2) {
    const dummy = new Node(0);
    let tail = dummy;
 
    while (true) {
      if (head1 === null) {
        tail.right = head2;
        break;
      }
 
      if (head2 === null) {
        tail.right = head1;
        break;
      }
 
      if (head1.data <= head2.data) {
        tail.right = head1;
        head1.left = tail;
        head1 = head1.right;
      } else {
        tail.right = head2;
        head2.left = tail;
        head2 = head2.right;
      }
 
      tail = tail.right;
    }
 
    const res = dummy.right;
    res.left = null;
    return res;
  }
 
  printList(node) {
    while (node !== null) {
      process.stdout.write(node.data+ " ");
      node = node.right;
    }
  }
}
 
const tree1 = new BinaryTree();
const tree2 = new BinaryTree();
 
tree1.insert(3);
tree1.insert(1);
tree1.insert(5);
 
tree2.insert(4);
tree2.insert(2);
tree2.insert(6);
 
const head1 = [null];
const head2 = [null];
 
tree1.convertIntoSortedDLL(tree1.root, head1);
tree2.convertIntoSortedDLL(tree2.root, head2);
 
const ans = tree1.mergeLinkedList(head1[0], head2[0]);
 
tree1.printList(ans);
 
// The code is contributed by Nidhi goel.

                    

Output
1 2 3 4 5 6 

Time Complexity: O(M+N), M is the size of the first tree and N is the size of the second tree
Auxiliary Space: O(max(H1,H2)), H1 is the height of the first tree and H2 is the height of the second tree



Last Updated : 31 Mar, 2023
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