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# Merge two BSTs with limited extra space

• Difficulty Level : Hard
• Last Updated : 26 Dec, 2022

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 ` `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<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<data<<``" "``; ` `                ``current1 = current1->right; ` `                ``push(&s2, current2); ` `                ``current2 = NULL; ` `            ``} ` `            ``else` `            ``{ ` `                ``cout<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` `#include`   `// 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

 ``

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 ` `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 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 res = ``new` `ArrayList();` `        ``Stack s1 = ``new` `Stack();` `        ``Stack s2 = ``new` `Stack();`   `        ``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 s1 = ``new` `Stack();` `        ``Stack s2 = ``new` `Stack();`   `        ``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