Merge two BSTs with constant extra space

Given two Binary Search Trees(BST), print the elements of both BSTs in sorted form. 
Note: Both the BSTs will not have any common element.

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

The idea is to use the fact the leftmost element (first in inorder traversal) of the tree is the least element in a BST. So we compute this value for both the trees and print the smaller one, now we delete this printed element from the respective tree and update it. Then we recursively call our function with the updated tree. We do this until one of the trees is exhausted. Now we simply print the inorder traversal of the other tree.

Below is the implementation of above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Structure of a BST Node
class Node {
public:
    int data;
    Node* left;
    Node* right;
    Node(int x)
    {
        data = x;
        left = right = NULL;
    }
};

// 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)
{
    // Base cases
    if (!root1 && !root2)
        return;

    // If the first tree is exhausted
    // simply print the inorder
    // traversal of the second tree
    if (!root1) {
        inorder(root2);
        return;
    }

    // If second tree is exhausted
    // simply print the inoreder
    // traversal of the first tree
    if (!root2) {
        inorder(root1);
        return;
    }

    // A temporary pointer currently
    // pointing to root of first tree
    Node* temp1 = root1;

    // previous pointer to store the
    // parent of temporary pointer
    Node* prev1 = NULL;

    // Traverse through the first tree until you reach
    // the leftmost element, which is the first element
    // of the tree in the inorder traversal.
    // This is the least element of the tree
    while (temp1->left) {
        prev1 = temp1;
        temp1 = temp1->left;
    }

    // Another temporary pointer currently
    // pointing to root of second tree
    Node* temp2 = root2;

    // Previous pointer to store the
    // parent of second temporary pointer
    Node* prev2 = NULL;

    // Traverse through the second tree until you reach
    // the leftmost element, which is the first element of
    // the tree in inorder traversal.
    // This is the least element of the tree.
    while (temp2->left) {
        prev2 = temp2;
        temp2 = temp2->left;
    }

    // Compare the least current least
    // elements of both the tree
    if (temp1->data <= temp2->data) {

        // If first tree's element is smaller print it
        cout << temp1->data << " ";

        // If the node has no parent, that
        // means this node is the root
        if (prev1 == NULL) {

            // Simply make the right
            // child of the root as new root
            merge(root1->right, root2);
        }

        // If node has a parent
        else {

            // As this node is the leftmost node,
            // it is certain that it will not have
            // a let child so we simply assign this
            // node's right pointer, which can be
            // either null or not, to its parent's left
            // pointer. This statement is
            // just doing the task of deleting the node

            prev1->left = temp1->right;

            // recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
    else {

        cout << temp2->data << " ";

        // If the node has no parent, that
        // means this node is the root
        if (prev2 == NULL) {

            // Simply make the right child
            // of root as new root
            merge(root1, root2->right);
        }

        // If node has a parent
        else {
            prev2->left = temp2->right;

            // Recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
}

// Driver Code
int main()
{
    Node *root1 = NULL, *root2 = NULL;
    root1 = new Node(3);
    root1->left = new Node(1);
    root1->right = new Node(5);
    root2 = new Node(4);
    root2->left = new Node(2);
    root2->right = new Node(6);

    // Print sorted nodes of both trees
    merge(root1, root2);

    return 0;
}

Java

// Java implementation of above approach
import java.util.*;

class GFG{

// Structure of a BST Node
static class Node
{
    int data;
    Node left;
    Node right;
};

static Node newNode(int num)
{
    Node temp = new Node();
    temp.data = num;
    temp.left = temp.right = null;
    return temp;
}

// 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)
{

    // Base cases
    if (root1 == null && root2 == null)
        return;

    // If the first tree is exhausted
    // simply print the inorder
    // traversal of the second tree
    if (root1 == null)
    {
        inorder(root2);
        return;
    }

    // If second tree is exhausted
    // simply print the inoreder
    // traversal of the first tree
    if (root2 == null)
    {
        inorder(root1);
        return;
    }

    // A temporary pointer currently
    // pointing to root of first tree
    Node temp1 = root1;

    // previous pointer to store the
    // parent of temporary pointer
    Node prev1 = null;

    // Traverse through the first tree
    // until you reach the leftmost element,
    // which is the first element of the tree
    // in the inorder traversal.
    // This is the least element of the tree
    while (temp1.left != null)
    {
        prev1 = temp1;
        temp1 = temp1.left;
    }

    // Another temporary pointer currently
    // pointing to root of second tree
    Node temp2 = root2;

    // Previous pointer to store the
    // parent of second temporary pointer
    Node prev2 = null;

    // Traverse through the second tree
    // until you reach the leftmost element,
    // which is the first element of
    // the tree in inorder traversal.
    // This is the least element of the tree.
    while (temp2.left != null)
    {
        prev2 = temp2;
        temp2 = temp2.left;
    }

    // Compare the least current least
    // elements of both the tree
    if (temp1.data <= temp2.data)
    {

        // If first tree's element is
        // smaller print it
        System.out.print(temp1.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev1 == null)
        {

            // Simply make the right
            // child of the root as new root
            merge(root1.right, root2);
        }

        // If node has a parent
        else
        {

            // As this node is the leftmost node,
            // it is certain that it will not have
            // a let child so we simply assign this
            // node's right pointer, which can be
            // either null or not, to its parent's left
            // pointer. This statement is
            // just doing the task of deleting the node
            prev1.left = temp1.right;

            // recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
    else
    {
        System.out.print(temp2.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev2 == null)
        {

            // Simply make the right child
            // of root as new root
            merge(root1, root2.right);
        }

        // If node has a parent
        else
        {
            prev2.left = temp2.right;

            // Recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
}

// Driver Code
public static void main(String args[])
{
    Node root1 = null, root2 = null;

    root1 = newNode(3);
    root1.left = newNode(1);
    root1.right = newNode(5);

    root2 = newNode(4);
    root2.left = newNode(2);
    root2.right = newNode(6);

    // Print sorted nodes of both trees
    merge(root1, root2);
}
}

// This code is contributed by ipg2016107

Python3

# Python3 implementation of above approach

# Node of the binary tree
class node:

    def __init__ (self, key):

        self.data = key
        self.left = None
        self.right = None

# A utility function to print
# Inorder traversal of a Binary Tree
def inorder(root):

    if (root != None):
        inorder(root.left)
        print(root.data, end = " ")
        inorder(root.right)

# The function to print data
# of two BSTs in sorted order
def merge(root1, root2):

    # Base cases
    if (not root1 and not root2):
        return

    # If the first tree is exhausted
    # simply print the inorder
    # traversal of the second tree
    if (not root1):
        inorder(root2)
        return

    # If second tree is exhausted
    # simply print the inoreder
    # traversal of the first tree
    if (not root2):
        inorder(root1)
        return

    # A temporary pointer currently
    # pointing to root of first tree
    temp1 = root1

    # previous pointer to store the
    # parent of temporary pointer
    prev1 = None

    # Traverse through the first tree
    # until you reach the leftmost
    # element, which is the first element
    # of the tree in the inorder traversal.
    # This is the least element of the tree
    while (temp1.left):
        prev1 = temp1
        temp1 = temp1.left

    # Another temporary pointer currently
    # pointing to root of second tree
    temp2 = root2

    # Previous pointer to store the
    # parent of second temporary pointer
    prev2 = None

    # Traverse through the second tree
    # until you reach the leftmost element,
    # which is the first element of the
    # tree in inorder traversal. This is
    # the least element of the tree.
    while (temp2.left):
        prev2 = temp2
        temp2 = temp2.left

    # Compare the least current least
    # elements of both the tree
    if (temp1.data <= temp2.data):

        # If first tree's element is
        # smaller print it
        print(temp1.data, end = " ")

        # If the node has no parent, that
        # means this node is the root
        if (prev1 == None):

            # Simply make the right
            # child of the root as new root
            merge(root1.right, root2)

        # If node has a parent
        else:

            # As this node is the leftmost node,
            # it is certain that it will not have
            # a let child so we simply assign this
            # node's right pointer, which can be
            # either null or not, to its parent's left
            # pointer. This statement is
            # just doing the task of deleting the node
            prev1.left = temp1.right

            # recursively call the merge
            # function with updated tree
            merge(root1, root2)
    else:

        print(temp2.data, end = " ")

        # If the node has no parent, that
        # means this node is the root
        if (prev2 == None):

            # Simply make the right child
            # of root as new root
            merge(root1, root2.right)

        # If node has a parent
        else:
            prev2.left = temp2.right

            # Recursively call the merge
            # function with updated tree
            merge(root1, root2)

# Driver Code
if __name__ == '__main__':

    root1 = None
    root2 = None
    root1 = node(3)
    root1.left = node(1)
    root1.right = node(5)
    root2 = node(4)
    root2.left = node(2)
    root2.right = node(6)

    # Print sorted nodes of both trees
    merge(root1, root2)

# This code is contributed by mohit kumar 29

C#

// C# implementation of above approach
using System;

// Structure of a BST Node
public class Node
{
    public int data;
    public Node left, right;

    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class GFG{

static Node root1;
static Node root2;

// A utility function to print
// Inorder traversal of a Binary Tree
static void inorder(Node root)
{
    if (root != null)
    {
        inorder(root.left);
        Console.WriteLine(root.data + " ");
        inorder(root.right);
    }
}

// The function to print data
// of two BSTs in sorted order
static void merge(Node root1, Node root2)
{

    // Base cases
    if (root1 == null && root2 == null)
    {
        return;
    }

    // If the first tree is exhausted
    // simply print the inorder traversal
    // of the second tree
    if (root1 == null)
    {
        inorder(root2);
        return;
    }

    // If second tree is exhausted
    // simply print the inoreder
    // traversal of the first tree
    if (root2 == null)
    {
        inorder(root1);
        return;
    }

    // A temporary pointer currently
    // pointing to root of first tree
    Node temp1 = root1;

    // previous pointer to store the
    // parent of temporary pointer
    Node prev1 = null;

    // Traverse through the first tree
    // until you reach the leftmost element,
    // which is the first element of the tree
    // in the inorder traversal.
    // This is the least element of the tree
    while (temp1.left != null)
    {
        prev1 = temp1;
        temp1 = temp1.left;
    }

    // Another temporary pointer currently
    // pointing to root of second tree
    Node temp2 = root2;

    // Previous pointer to store the
    // parent of second temporary pointer
    Node prev2 = null;

    // Traverse through the second tree until
    // you reach the leftmost element, which
    // is the first element of the tree in
    // inorder traversal. This is the least
    // element of the tree.
    while (temp2.left != null)
    {
        prev2 = temp2;
        temp2 = temp2.left;
    }

    // Compare the least current least
    // elements of both the tree
    if (temp1.data <= temp2.data)
    {

        // If first tree's element is
        // smaller print it
        Console.Write(temp1.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev1 == null)
        {

            // Simply make the right
            // child of the root as new root
            merge(root1.right, root2);
        }

        // If node has a parent
        else
        {

            // As this node is the leftmost node,
            // it is certain that it will not have
            // a let child so we simply assign this
            // node's right pointer, which can be
            // either null or not, to its parent's
            // left pointer. This statement is just
            // doing the task of deleting the node
            prev1.left = temp1.right;

            // Recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
    else
    {
        Console.Write(temp2.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev2 == null)
        {

            // Simply make the right child
            // of root as new root
            merge(root1, root2.right);
        }

        // If node has a parent
        else
        {
            prev2.left = temp2.right;

            // Recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
}

// Driver Code
static public void Main()
{
    GFG.root1 = new Node(3);
    GFG.root1.left = new Node(1);
    GFG.root1.right = new Node(5);

    GFG.root2 = new Node(4);
    GFG.root2.left = new Node(2);
    GFG.root2.right = new Node(6);

    // Print sorted nodes of both trees
    merge(root1, root2);
}
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>

// Javascript implementation of above approach

// Structure of a BST Node
class Node
{
    constructor(num)
    {
        this.data=num;
        this.left=this.right=null;
    }
}

// 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)
{
    // Base cases
    if (root1 == null && root2 == null)
        return;

    // If the first tree is exhausted
    // simply print the inorder
    // traversal of the second tree
    if (root1 == null)
    {
        inorder(root2);
        return;
    }

    // If second tree is exhausted
    // simply print the inoreder
    // traversal of the first tree
    if (root2 == null)
    {
        inorder(root1);
        return;
    }

    // A temporary pointer currently
    // pointing to root of first tree
    let temp1 = root1;

    // previous pointer to store the
    // parent of temporary pointer
    let prev1 = null;

    // Traverse through the first tree
    // until you reach the leftmost element,
    // which is the first element of the tree
    // in the inorder traversal.
    // This is the least element of the tree
    while (temp1.left != null)
    {
        prev1 = temp1;
        temp1 = temp1.left;
    }

    // Another temporary pointer currently
    // pointing to root of second tree
    let temp2 = root2;

    // Previous pointer to store the
    // parent of second temporary pointer
    let prev2 = null;

    // Traverse through the second tree
    // until you reach the leftmost element,
    // which is the first element of
    // the tree in inorder traversal.
    // This is the least element of the tree.
    while (temp2.left != null)
    {
        prev2 = temp2;
        temp2 = temp2.left;
    }

    // Compare the least current least
    // elements of both the tree
    if (temp1.data <= temp2.data)
    {

        // If first tree's element is
        // smaller print it
        document.write(temp1.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev1 == null)
        {

            // Simply make the right
            // child of the root as new root
            merge(root1.right, root2);
        }

        // If node has a parent
        else
        {

            // As this node is the leftmost node,
            // it is certain that it will not have
            // a let child so we simply assign this
            // node's right pointer, which can be
            // either null or not, to its parent's left
            // pointer. This statement is
            // just doing the task of deleting the node
            prev1.left = temp1.right;

            // recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
    else
    {
        document.write(temp2.data + " ");

        // If the node has no parent, that
        // means this node is the root
        if (prev2 == null)
        {

            // Simply make the right child
            // of root as new root
            merge(root1, root2.right);
        }

        // If node has a parent
        else
        {
            prev2.left = temp2.right;

            // Recursively call the merge
            // function with updated tree
            merge(root1, root2);
        }
    }
}

// Driver Code
let root1 = null, root2 = null;

    root1 = new Node(3);
    root1.left = new Node(1);
    root1.right = new Node(5);

    root2 = new Node(4);
    root2.left = new Node(2);
    root2.right = new Node(6);

    // Print sorted nodes of both trees
    merge(root1, root2);


// This code is contributed by unknown2108
</script>

Output

1 2 3 4 5 6 

Time Complexity: O((M+N)(h1+h2)), where M and N are the number of nodes of the two trees and, h1 and h2 are the heights of tree respectively. 
Auxiliary Space: O(N)

Method 2: Using Morris Traversal of the trees

The above solution uses recursion and hence require O(N) auxiliary space. We can use the concept of Morris Traversal to improve the space complexity of the algorithm by eliminating the usage of stacks. Morris traversal is a method of traversing a binary tree without using recursion and with constant (O(1)) extra space.

The idea is to traverse to the smallest data node of both trees using Morris traversal and compare the nodes on both trees. We use the smaller one to add to the answer array and move it to the next node in sorted order. We know that inorder traversal of Binary Search Trees is in the sorted order. Thus we can move to the smallest node and continue moving in the inorder order to get to the next higher node.

Algorithm:

1. Apply Morris traversal on root1:
   • Check if root has a left node and if it has a left node.
     • Go to the rightmost node of the left node.
     • On the right most node of the left, assign its right pointer to root.
     • Make left = root->left and make root->left = NULL
     • Move root to left.
   • Repeat till root has a left node
   • If it doesn’t have a left node, that means root is the smallest node, thus break root1’s Morris traversal.
2. Apply same Morris traversal for root2.
3. Now root1 and root2 are at the smallest nodes of their respective trees. Compare root1 data with root2 data:
   • If root1 data is smaller than or equal to root2 data: add root1 data to our answer array and move root1 to its right.
   • Else add root2 data to our answer array and move root2 to its right.
4. If in step 3, root1 or root2 is NULL, that means that tree is exhausted and we have added all nodes to answer array, so we just add the remaining tree data to answer array without comparison and move it to its right.
5. Repeat steps 1 to 4 until both trees are exhausted.

C++

#include <bits/stdc++.h>
using namespace std;

// Structure of a BST Node
class Node {
public:
   int data;
   Node *left, *right;

   // Constructor
   Node(int data){
       this->data = data;
       this->left = NULL;
       this->right = NULL;
   }
};

void mergeBSTs(Node*, Node*);

int main(){
   /* Let us create the following tree as first tree
          3
         / \
         1 5
   */

   Node* 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
   */

   Node* root2 = new Node(4);
   root2->left = new Node(2);
   root2->right = new Node(6);

   // Merging the BSTs
   mergeBSTs(root1, root2);
}



void mergeBSTs(Node* root1, Node* root2){
   // We run this loop until both trees are completely
   // exhausted Even if one of the trees is still left, we
   // run this loop
   while (root1 || root2) {
       // Morris traversal of the first tree
       while (root1) { // This check is to ensure that if
                       // root1 is already exhausted we skip root1

            // If root has a left node, we go to the
            // rightmost child of the left node and assign
               // root to the right of the rightmost node
           if (root1->left) {
               Node* left = root1->left;

               // Moving to the rightmost node of left
               while (left->right)
                   left = left->right;

               // Assign root to right of rightmost node
               left->right = root1;

               // Make root's left to NULL and move root to left
               left = root1->left;
               root1->left = NULL;
               root1 = left;
           }
           else break; // If root doesn't have a left node, that means
              // we're already on the left most (smallest) node
       }

       // Morris traversal of the second tree
       while (root2) { // This check is to ensure that if
                       // root2 is already exhausted we skip root2

           // If root has a left node, we go to the
           // rightmost child of the left node and assign
           // root to the right of the rightmost node
           if (root2->left) {
               Node* left = root2->left;

               // Moving to the rightmost node of left
               while (left->right)
                   left = left->right;

               // Assign root to right of rightmost node
               left->right = root2;

               // Make root's left to NULL and move root to left
               left = root2->left;
               root2->left = NULL;
               root2 = left;
           }
           else break; // If root doesn't have a left node, that means
                      // we're already on the left most (smallest) node
       }

       // Here root1 and root2 are smallest nodes in both trees
       if (root1 && root2) {
           // Compare both nodes' data
           if (root1->data <= root2->data) {
               cout << root1->data << " "; // Add smaller one to ans array
               root1 = root1->right; // Move smaller one to right
           }
           else {
               cout << root2->data << " "; // Add smaller one to ans array
               root2 = root2->right; // Move smaller one to right
           }
       }
       else if (root1) { // If root2 has exhausted and only root1 remains
           cout << root1->data << " "; // Add it to ans array
           root1 = root1->right; // Move it to right
       }
       else if (root2) { // If root2 has exhausted and only root2 remains
           cout << root2->data << " "; // Add it to ans array
           root2 = root2->right; // Move it to right
       }
   }
}

Java

import java.util.*;

class GFG{

// Structure of a BST Node
static class Node {
   int data;
   Node left, right;

   // Constructor
   Node(int data){
       this.data = data;
       this.left = null;
       this.right = null;
   }
};

public static void main(String[] args){
   /* Let us create the following tree as first tree
          3
         / \
         1 5
   */

   Node 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
   */

   Node root2 = new Node(4);
   root2.left = new Node(2);
   root2.right = new Node(6);

   // Merging the BSTs
   mergeBSTs(root1, root2);
}



static void mergeBSTs(Node root1, Node root2)
{

   // We run this loop until both trees are completely
   // exhausted Even if one of the trees is still left, we
   // run this loop
   while (root1 != null || root2 != null)
   {

       // Morris traversal of the first tree
       while (root1 != null)
       {

         // This check is to ensure that if
         // root1 is already exhausted we skip root1

            // If root has a left node, we go to the
            // rightmost child of the left node and assign
               // root to the right of the rightmost node
           if (root1.left != null)
           {
               Node left = root1.left;

               // Moving to the rightmost node of left
               while (left.right != null)
                   left = left.right;

               // Assign root to right of rightmost node
               left.right = root1;

               // Make root's left to null and move root to left
               left = root1.left;
               root1.left = null;
               root1 = left;
           }
           else break; // If root doesn't have a left node, that means
              // we're already on the left most (smallest) node
       }

       // Morris traversal of the second tree
       while (root2!=null) { // This check is to ensure that if
                       // root2 is already exhausted we skip root2

           // If root has a left node, we go to the
           // rightmost child of the left node and assign
           // root to the right of the rightmost node
           if (root2.left!=null) {
               Node left = root2.left;

               // Moving to the rightmost node of left
               while (left.right!=null)
                   left = left.right;

               // Assign root to right of rightmost node
               left.right = root2;

               // Make root's left to null and move root to left
               left = root2.left;
               root2.left = null;
               root2 = left;
           }
           else break; // If root doesn't have a left node, that means
                      // we're already on the left most (smallest) node
       }

       // Here root1 and root2 are smallest nodes in both trees
       if (root1!=null && root2!=null) {
           // Compare both nodes' data
           if (root1.data <= root2.data) {
               System.out.print(root1.data+ " "); // Add smaller one to ans array
               root1 = root1.right; // Move smaller one to right
           }
           else {
               System.out.print(root2.data+ " "); // Add smaller one to ans array
               root2 = root2.right; // Move smaller one to right
           }
       }
       else if (root1 != null) { // If root2 has exhausted and only root1 remains
           System.out.print(root1.data+ " "); // Add it to ans array
           root1 = root1.right; // Move it to right
       }
       else if (root2 != null) { // If root2 has exhausted and only root2 remains
           System.out.print(root2.data+ " "); // Add it to ans array
           root2 = root2.right; // Move it to right
       }
   }
}
}

// This code is contributed by Rajput-Ji

Python3

# Python code for the above approach
class node:
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None

def mergeBSTs(root1, root2):
    # We run this loop until both trees are completely
    # exhausted Even if one of the trees is still left, we
    # run this loop
    while (root1 or root2):
        # Morris traversal of the first tree
        while(root1):
            # This check is to ensure that if
            # root1 is already exhausted we skip root1

            # If root has a left node, we go to the
            # rightmost child of the left node and assign
            # root to the right of the rightmost node
            if(root1.left):
                left = root1.left
                # Moving to the rightmost node of left
                while (left.right):
                    left = left.right
                # Assign root to right of rightmost node
                left.right = root1
                # Make root's left to null and move root to left
                left = root1.left
                root1.left = None
                root1 = left
            else:
                # If root doesn't have a left node, that means
                # we're already on the left most (smallest) node
                break

        # Morris traversal of the second tree
        while(root2):  # This check is to ensure that if
                      # root2 is already exhausted we skip root2

            # If root has a left node, we go to the
            # rightmost child of the left node and assign
            # root to the right of the rightmost node
            if (root2.left):
                left = root2.left
                # Moving to the rightmost node of left
                while(left.right):
                    left = left.right
                # Assign root to right of rightmost node
                left.right = root2
                # Make root's left to null and move root to left
                left = root2.left
                root2.left = None
                root2 = left
            else:
                break

        # Here root1 and root2 are smallest nodes in both trees
        if(root1 and root2):
            # Compare both nodes' data
            if(root1.data <= root2.data):
                print(root1.data, end=" ")  # Add smaller one to ans array
                root1 = root1.right  # Move smaller one to right
            else:
                print(root2.data, end=" ")  # Add smaller one to ans array
                root2 = root2.right  # Move smaller one to right
        elif(root1):  # If root2 has exhausted and only root1 remains
            print(root1.data, end=" ")  # Add it to ans array
            root1 = root1.right  # Move it to right
        elif(root2):  # If root2 has exhausted and only root2 remains
            print(root2.data, end=" ")  # Add it to ans array
            root2 = root2.right  # Move it to right

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 first tree
    #       4
    #      / \
    #      2 6

    root2 = node(4)
    root2.left = node(2)
    root2.right = node(6)

    # Merging the BSTs
    mergeBSTs(root1, root2)

# This code is contributed by lokesh (lokeshmvs21).

C#

using System;

public class GFG {

  // Structure of a BST Node
  public class Node {
    public int data;
    public Node left, right;

    // Constructor
    public Node(int data) {
      this.data = data;
      this.left = null;
      this.right = null;
    }
  };

  public static void Main(String[] args) {
    /*
         * Let us create the following tree as first tree 3 / \ 1 5
         */

    Node 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
         */

    Node root2 = new Node(4);
    root2.left = new Node(2);
    root2.right = new Node(6);

    // Merging the BSTs
    mergeBSTs(root1, root2);
  }

  static void mergeBSTs(Node root1, Node root2) {

    // We run this loop until both trees are completely
    // exhausted Even if one of the trees is still left, we
    // run this loop
    while (root1 != null || root2 != null) {

      // Morris traversal of the first tree
      while (root1 != null) {

        // This check is to ensure that if
        // root1 is already exhausted we skip root1

        // If root has a left node, we go to the
        // rightmost child of the left node and assign
        // root to the right of the rightmost node
        if (root1.left != null) {
          Node left = root1.left;

          // Moving to the rightmost node of left
          while (left.right != null)
            left = left.right;

          // Assign root to right of rightmost node
          left.right = root1;

          // Make root's left to null and move root to left
          left = root1.left;
          root1.left = null;
          root1 = left;
        } else
          break; // If root doesn't have a left node, that means
        // we're already on the left most (smallest) node
      }

      // Morris traversal of the second tree
      while (root2 != null)
      {
        // This check is to ensure that if
        // root2 is already exhausted we skip root2

        // If root has a left node, we go to the
        // rightmost child of the left node and assign
        // root to the right of the rightmost node
        if (root2.left != null) {
          Node left = root2.left;

          // Moving to the rightmost node of left
          while (left.right != null)
            left = left.right;

          // Assign root to right of rightmost node
          left.right = root2;

          // Make root's left to null and move root to left
          left = root2.left;
          root2.left = null;
          root2 = left;
        } else
          break; // If root doesn't have a left node, that means
        // we're already on the left most (smallest) node
      }

      // Here root1 and root2 are smallest nodes in both trees
      if (root1 != null && root2 != null)
      {

        // Compare both nodes' data
        if (root1.data <= root2.data) {
          Console.Write(root1.data + " "); // Add smaller one to ans array
          root1 = root1.right; // Move smaller one to right
        } else {
          Console.Write(root2.data + " "); // Add smaller one to ans array
          root2 = root2.right; // Move smaller one to right
        }
      } else if (root1 != null)
      {

        // If root2 has exhausted and only root1 remains
        Console.Write(root1.data + " "); // Add it to ans array
        root1 = root1.right; // Move it to right
      } else if (root2 != null)
      {

        // If root2 has exhausted and only root2 remains
        Console.Write(root2.data + " "); // Add it to ans array
        root2 = root2.right; // Move it to right
      }
    }
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
    // Structure of a BST Node
     class Node
     {

        // Constructor
        constructor(data) {
            this.data = data;
            this.left = null;
            this.right = null;
        }
}

    function mergeBSTs(root1, root2)
    {

        // We run this loop until both trees are completely
        // exhausted Even if one of the trees is still left, we
        // run this loop
        while (root1 != null || root2 != null) {

            // Morris traversal of the first tree
            while (root1 != null) {

                // This check is to ensure that if
                // root1 is already exhausted we skip root1

                // If root has a left node, we go to the
                // rightmost child of the left node and assign
                // root to the right of the rightmost node
                if (root1.left != null) {
            var left = root1.left;

                    // Moving to the rightmost node of left
                    while (left.right != null)
                        left = left.right;

                    // Assign root to right of rightmost node
                    left.right = root1;

                    // Make root's left to null and move root to left
                    left = root1.left;
                    root1.left = null;
                    root1 = left;
                } else
                    break; // If root doesn't have a left node, that means
                // we're already on the left most (smallest) node
            }

            // Morris traversal of the second tree
            while (root2 != null) { // This check is to ensure that if
                // root2 is already exhausted we skip root2

                // If root has a left node, we go to the
                // rightmost child of the left node and assign
                // root to the right of the rightmost node
                if (root2.left != null) {
            var left = root2.left;

                    // Moving to the rightmost node of left
                    while (left.right != null)
                        left = left.right;

                    // Assign root to right of rightmost node
                    left.right = root2;

                    // Make root's left to null and move root to left
                    left = root2.left;
                    root2.left = null;
                    root2 = left;
                } else
                    break; // If root doesn't have a left node, that means
                            // we're already on the left most (smallest) node
            }

            // Here root1 and root2 are smallest nodes in both trees
            if (root1 != null && root2 != null) {
                // Compare both nodes' data
                if (root1.data <= root2.data) {
                    document.write(root1.data + " "); // Add smaller one to ans array
                    root1 = root1.right; // Move smaller one to right
                } else {
                    document.write(root2.data + " "); // Add smaller one to ans array
                    root2 = root2.right; // Move smaller one to right
                }
            } else if (root1 != null) { // If root2 has exhausted and only root1 remains
                document.write(root1.data + " "); // Add it to ans array
                root1 = root1.right; // Move it to right
            } else if (root2 != null) { // If root2 has exhausted and only root2 remains
                document.write(root2.data + " "); // Add it to ans array
                root2 = root2.right; // Move it to right
            }
        }
    }
/*
         * Let us create the following tree as first tree
          3
         / \
         1 5
         */

        var 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
         */

        var root2 = new Node(4);
        root2.left = new Node(2);
        root2.right = new Node(6);

        // Merging the BSTs
        mergeBSTs(root1, root2);

// This code is contributed by Rajput-Ji
</script>

Output

1 2 3 4 5 6 

Time Complexity: O(m + n)

Auxiliary Space Complexity: O(1).

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