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Merge two BSTs with constant extra space
• Difficulty Level : Hard
• Last Updated : 08 Nov, 2020

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 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

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(1)

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