# Merge two BSTs with limited extra space

• Difficulty Level : Hard
• Last Updated : 12 Apr, 2022

Given two Binary Search Trees(BST), print the elements of both BSTs in sorted form. The expected time complexity is O(m+n) where m is the number of nodes in the first tree and n is the number of nodes in the second tree. The maximum allowed auxiliary space is O(height of the first tree + height of the second tree).
Examples:

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

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

A similar question has been discussed earlier. Let us first discuss the already discussed methods of the previous post which was for Balanced BSTs. Method 1 can be applied here also, but the time complexity will be O(n^2) in the worst case. Method 2 can also be applied here, but the extra space required will be O(n) which violates the constraint given in this question. Method 3 can be applied here but step 3 of method 3 can’t be done in O(n) for an unbalanced BST.
Thanks to Kumar for suggesting the following solution.
The idea is to use iterative inorder traversal. We use two auxiliary stacks for two BSTs. Since we need to print the elements in the sorted form, whenever we get a smaller element from any of the trees, we print it. If the element is greater, then we push it back to stack for the next iteration.

## 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)
Auxiliary Space: O(height of the first tree + height of the second tree)

Another simpler version of solving the problem using two stacks is given below :-

In this method I am using inbuilt stack present in the STL library so as to get rid of the implementation of the stack part of code that has done in the previous implementation.

Step 1 :- Consider two stacks s1 and s2 which stores the element of the two trees.

Step 2 :- Store the left view value of a tree1 in s1 and of tree2 in s2.

Step 3 :- Compare the top values present in the stack and push the value accordingly in the result vector.

Case 1 :- 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

Case 2 :- 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

Step 4 :- Continue step 3 until both the stacks get empty.

Below is the implementation of the above logic :-

## 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)`
Output
`1 2 3 4 5 6 `

Time Complexity: O(m+n)
Auxiliary Space: O(height of the first tree + height of the second tree)