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# Symmetric Tree (Mirror Image of itself)

Given a binary tree, check whether it is a mirror of itself.

For example, this binary tree is symmetric:

```     1
/   \
2     2
/ \   / \
3   4 4   3

But the following is not:
1
/ \
2   2
\   \
3    3```
Recommended Practice

The idea is to write a recursive function isMirror() that takes two trees as an argument and returns true if trees are the mirror and false if trees are not mirrored. The isMirror() function recursively checks two roots and subtrees under the root.

Below is the implementation of the above algorithm.

## C

 `// C program to check if a given Binary Tree is symmetric``// or not``#include ``#include ``#include ` `// A Binary Tree Node``typedef` `struct` `Node {``    ``int` `key;``    ``struct` `Node *left, *right;``} Node;` `// Utility function to create new Node``Node* newNode(``int` `key)``{``    ``Node* temp = (Node *)``malloc``(``sizeof``(Node));``    ``temp->key = key;``    ``temp->left = temp->right = NULL;``    ``return` `(temp);``}` `// Returns true if trees with roots as root1 and root2 are``// mirror``bool` `isMirror(Node* root1, Node* root2)``{``    ``// If both trees are empty, then they are mirror images``    ``if` `(root1 == NULL && root2 == NULL)``        ``return` `true``;` `    ``// For two trees to be mirror images, the following``    ``// three conditions must be true``    ``// 1.) Their root node's key must be same``    ``// 2.) left subtree of left tree and right subtree of``    ``// right tree have to be mirror images``    ``// 3.) right subtree of left tree and left subtree of``    ``// right tree have to be mirror images``    ``if` `(root1 && root2 && root1->key == root2->key)``        ``return` `isMirror(root1->left, root2->right)``               ``&& isMirror(root1->right, root2->left);` `    ``// if none of above conditions is true then root1``    ``// and root2 are not mirror images``    ``return` `false``;``}` `// Returns true if a tree is symmetric i.e. mirror image of``// itself``bool` `isSymmetric(Node* root)``{``    ``// Check if tree is mirror of itself``    ``return` `isMirror(root, root);``}` `// Driver code``int` `main()``{``    ``// Let us construct the Tree shown in the above figure``    ``Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(2);``    ``root->left->left = newNode(3);``    ``root->left->right = newNode(4);``    ``root->right->left = newNode(4);``    ``root->right->right = newNode(3);` `    ``if` `(isSymmetric(root))``        ``printf``(``"Symmetric"``);``    ``else``        ``printf``(``"Not symmetric"``);``    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta`

## C++14

 `// C++ program to check if a given Binary Tree is symmetric``// or not``#include ``using` `namespace` `std;` `// A Binary Tree Node``struct` `Node {``    ``int` `key;``    ``struct` `Node *left, *right;``};` `// Utility function to create new Node``Node* newNode(``int` `key)``{``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``temp->left = temp->right = NULL;``    ``return` `(temp);``}` `// Returns true if trees with roots as root1 and root2 are``// mirror``bool` `isMirror(``struct` `Node* root1, ``struct` `Node* root2)``{``    ``// If both trees are empty, then they are mirror images``    ``if` `(root1 == NULL && root2 == NULL)``        ``return` `true``;` `    ``// For two trees to be mirror images, the following``    ``// three conditions must be true``    ``// 1.) Their root node's key must be same``    ``// 2.) left subtree of left tree and right subtree of``    ``// right tree have to be mirror images``    ``// 3.) right subtree of left tree and left subtree of``    ``// right tree have to be mirror images``    ``if` `(root1 && root2 && root1->key == root2->key)``        ``return` `isMirror(root1->left, root2->right)``               ``&& isMirror(root1->right, root2->left);` `    ``// if none of above conditions is true then root1``    ``// and root2 are not mirror images``    ``return` `false``;``}` `// Returns true if a tree is symmetric i.e. mirror image of itself``bool` `isSymmetric(``struct` `Node* root)``{``    ``// Check if tree is mirror of itself``    ``return` `isMirror(root, root);``}` `// Driver code``int` `main()``{``    ``// Let us construct the Tree shown in the above figure``    ``Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(2);``    ``root->left->left = newNode(3);``    ``root->left->right = newNode(4);``    ``root->right->left = newNode(4);``    ``root->right->right = newNode(3);` `    ``if` `(isSymmetric(root))``        ``cout << ``"Symmetric"``;``    ``else``        ``cout << ``"Not symmetric"``;``    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta`

## Java

 `// Java program to check is binary tree is symmetric or not``class` `Node {``    ``int` `key;``    ``Node left, right;``    ``Node(``int` `item)``    ``{``        ``key = item;``        ``left = right = ``null``;``    ``}``}` `class` `BinaryTree {``    ``Node root;` `    ``// returns true if trees with roots as root1 and``    ``// root2 are mirror``    ``boolean` `isMirror(Node node1, Node node2)``    ``{``        ``// if both trees are empty, then they are mirror image``        ``if` `(node1 == ``null` `&& node2 == ``null``)``            ``return` `true``;` `        ``// For two trees to be mirror images, the following``        ``// three conditions must be true``        ``// 1.) Their root node's key must be same``        ``// 2.) left subtree of left tree and right subtree``        ``// of right tree have to be mirror images``        ``// 3.) right subtree of left tree and left subtree``        ``// of right tree have to be mirror images``        ``if` `(node1 != ``null` `&& node2 != ``null``            ``&& node1.key == node2.key)``            ``return` `(isMirror(node1.left, node2.right)``                    ``&& isMirror(node1.right, node2.left));` `        ``// if none of the above conditions is true then``        ``// root1 and root2 are not mirror images``        ``return` `false``;``    ``}` `    ``// returns true if the tree is symmetric i.e``    ``// mirror image of itself``    ``boolean` `isSymmetric()``    ``{``        ``// check if tree is mirror of itself``        ``return` `isMirror(root, root);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(``1``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``2``);``        ``tree.root.left.left = ``new` `Node(``3``);``        ``tree.root.left.right = ``new` `Node(``4``);``        ``tree.root.right.left = ``new` `Node(``4``);``        ``tree.root.right.right = ``new` `Node(``3``);``        ``boolean` `output = tree.isSymmetric();``        ``if` `(output == ``true``)``            ``System.out.println(``"Symmetric"``);``        ``else``            ``System.out.println(``"Not symmetric"``);``    ``}``}` `// This code is contributed by Sania Kumari Gupta`

## Python3

 `# Python program to check if a``# given Binary Tree is symmetric or not` `# Node structure`  `class` `Node:` `    ``# Utility function to create new node``    ``def` `__init__(``self``, key):``        ``self``.key ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Returns True if trees``#with roots as root1 and root 2  are mirror`  `def` `isMirror(root1, root2):``    ``# If both trees are empty, then they are mirror images``    ``if` `root1 ``is` `None` `and` `root2 ``is` `None``:``        ``return` `True` `    ``""" For two trees to be mirror images,``        ``the following three conditions must be true``        ``1 - Their root node's key must be same``        ``2 - left subtree of left tree and right subtree``          ``of the right tree have to be mirror images``        ``3 - right subtree of left tree and left subtree``           ``of right tree have to be mirror images``    ``"""``    ``if` `(root1 ``is` `not` `None` `and` `root2 ``is` `not` `None``):``        ``if` `root1.key ``=``=` `root2.key:``            ``return` `(isMirror(root1.left, root2.right)``and``                    ``isMirror(root1.right, root2.left))` `    ``# If none of the above conditions is true then root1``    ``# and root2 are not mirror images``    ``return` `False`  `def` `isSymmetric(root):` `    ``# Check if tree is mirror of itself``    ``return` `isMirror(root, root)`  `# Driver Code``# Let's construct the tree show in the above figure``root ``=` `Node(``1``)``root.left ``=` `Node(``2``)``root.right ``=` `Node(``2``)``root.left.left ``=` `Node(``3``)``root.left.right ``=` `Node(``4``)``root.right.left ``=` `Node(``4``)``root.right.right ``=` `Node(``3``)``print` `(``"Symmetric"` `if` `isSymmetric(root) ``=``=` `True` `else` `"Not symmetric"``)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## Javascript

 ``

## C#

 `// C# program to check is binary``// tree is symmetric or not``using` `System;` `class` `Node {``    ``public` `int` `key;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``key = item;``        ``left = right = ``null``;``    ``}``}` `class` `GFG {``    ``Node root;` `    ``// returns true if trees with roots``    ``// as root1 and root2 are mirror``    ``Boolean isMirror(Node node1, Node node2)``    ``{``        ``// if both trees are empty,``        ``// then they are mirror image``        ``if` `(node1 == ``null` `&& node2 == ``null``)``            ``return` `true``;` `        ``// For two trees to be mirror images,``        ``// the following three conditions must be true``        ``// 1 - Their root node's key must be same``        ``// 2 - left subtree of left tree and right``        ``// subtree of right tree have to be mirror images``        ``// 3 - right subtree of left tree and left subtree``        ``// of right tree have to be mirror images``        ``if` `(node1 != ``null` `&& node2 != ``null``            ``&& node1.key == node2.key)``            ``return` `(isMirror(node1.left, node2.right)``                    ``&& isMirror(node1.right, node2.left));` `        ``// if none of the above conditions``        ``// is true then root1 and root2 are``        ``// mirror images``        ``return` `false``;``    ``}` `    ``// returns true if the tree is symmetric``    ``// i.e mirror image of itself``    ``Boolean isSymmetric()``    ``{``        ``// check if tree is mirror of itself``        ``return` `isMirror(root, root);``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main(String[] args)``    ``{``        ``GFG tree = ``new` `GFG();``        ``tree.root = ``new` `Node(1);``        ``tree.root.left = ``new` `Node(2);``        ``tree.root.right = ``new` `Node(2);``        ``tree.root.left.left = ``new` `Node(3);``        ``tree.root.left.right = ``new` `Node(4);``        ``tree.root.right.left = ``new` `Node(4);``        ``tree.root.right.right = ``new` `Node(3);``        ``Boolean output = tree.isSymmetric();``        ``if` `(output == ``true``)``            ``Console.WriteLine(``"Symmetric"``);``        ``else``            ``Console.WriteLine(``"Not symmetric"``);``    ``}``}` `// This code is contributed by Arnab Kundu`

Output

`Symmetric`

Time Complexity: O(N)
Auxiliary Space: O(h) where h is the maximum height of the tree

Iterative Approach:

Algorithm for checking whether a binary tree is a mirror of itself using an iterative approach and a stack:

1. Create a stack and push the root node onto it twice.
2. While the stack is not empty, repeat the following steps:
a. Pop two nodes from the stack, say node1 and node2.
b. If both node1 and node2 are null, continue to the next iteration.
c. If one of the nodes is null and the other is not, return false as it is not a mirror.
d. If both nodes are not null, compare their values. If they are not equal, return false.
e. Push the left child of node1 and the right child of node2 onto the stack.
f. Push the right child of node1 and the left child of node2 onto the stack.
3. If the loop completes successfully without returning false, return true as it is a mirror.

## C++

 `// C++ program to check if a given Binary Tree is symmetric``// or not``#include ``using` `namespace` `std;` `// A Binary Tree Node``struct` `Node {``    ``int` `key;``    ``struct` `Node *left, *right;``};` `// Utility function to create new Node``Node* newNode(``int` `key)``{``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``temp->left = temp->right = NULL;``    ``return` `(temp);``}` `// Returns true if a tree is symmetric i.e. mirror image of``// itself``bool` `isSymmetric(Node* root){``    ``// If the root is null, then the binary tree is``    ``// symmetric.``    ``if` `(root == NULL) {``        ``return` `true``;``    ``}` `    ``// Create a stack to store the left and right subtrees``    ``// of the root.``    ``stack stack;``    ``stack.push(root->left);``    ``stack.push(root->right);` `    ``// Continue the loop until the stack is empty.``    ``while` `(!stack.empty()) {``        ``// Pop the left and right subtrees from the stack.``        ``Node* node1 = stack.top();``          ``stack.pop();``        ``Node* node2 = stack.top();``          ``stack.pop();` `        ``// If both nodes are null, continue the loop.``        ``if` `(node1 == NULL && node2 == NULL) {``            ``continue``;``        ``}` `        ``// If one of the nodes is null, the binary tree is``        ``// not symmetric.``        ``if` `(node1 == NULL || node2 == NULL) {``            ``return` `false``;``        ``}` `        ``// If the values of the nodes are not equal, the``        ``// binary tree is not symmetric.``        ``if` `(node1->key != node2->key) {``            ``return` `false``;``        ``}` `        ``// Push the left and right subtrees of the left and``        ``// right nodes onto the stack in the opposite order.``        ``stack.push(node1->left);``        ``stack.push(node2->right);``        ``stack.push(node1->right);``        ``stack.push(node2->left);``    ``}` `    ``// If the loop completes, the binary tree is symmetric.``    ``return` `true``;``}` `// Driver code``int` `main()``{``    ``// Let us construct the Tree shown in the above figure``    ``Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(2);``    ``root->left->left = newNode(3);``    ``root->left->right = newNode(4);``    ``root->right->left = newNode(4);``    ``root->right->right = newNode(3);` `    ``if` `(isSymmetric(root))``        ``cout << ``"Symmetric"``;``    ``else``        ``cout << ``"Not symmetric"``;``    ``return` `0;``}` `// This code is contributed by sramshyam`

## Java

 `// Java program to check if a given Binary Tree is symmetric``// or not``import` `java.util.*;` `// A Binary Tree Node``class` `Node {``  ``int` `key;``  ``Node left, right;``  ``// Constructor``  ``Node(``int` `item)``  ``{``    ``key = item;``    ``left = right = ``null``;``  ``}``}` `public` `class` `GFG``{``  ` `  ``// Returns true if a tree is symmetric i.e. mirror image``  ``// of itself``  ``static` `boolean` `isSymmetric(Node root)``  ``{``    ` `    ``// If the root is null, then the binary tree is``    ``// symmetric.``    ``if` `(root == ``null``) {``      ``return` `true``;``    ``}``    ` `    ``// Create a stack to store the left and right``    ``// subtrees``    ``// of the root.``    ``Stack stack = ``new` `Stack<>();``    ``stack.push(root.left);``    ``stack.push(root.right);` `    ``// Continue the loop until the stack is empty.``    ``while` `(!stack.empty()) {``      ``// Pop the left and right subtrees from the``      ``// stack.``      ``Node node1 = stack.pop();``      ``Node node2 = stack.pop();` `      ``// If both nodes are null, continue the loop.``      ``if` `(node1 == ``null` `&& node2 == ``null``) {``        ``continue``;``      ``}` `      ``// If one of the nodes is null, the binary tree``      ``// is not symmetric.``      ``if` `(node1 == ``null` `|| node2 == ``null``) {``        ``return` `false``;``      ``}` `      ``// If the values of the nodes are not equal, the``      ``// binary tree is not symmetric.``      ``if` `(node1.key != node2.key) {``        ``return` `false``;``      ``}` `      ``// Push the left and right subtrees of the left``      ``// and right nodes onto the stack in the``      ``// opposite order.``      ``stack.push(node1.left);``      ``stack.push(node2.right);``      ``stack.push(node1.right);``      ``stack.push(node2.left);``    ``}` `    ``// If the loop completes, the binary tree is``    ``// symmetric.``    ``return` `true``;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``// Let us construct the Tree shown in the above``    ``// figure``    ``Node root = ``new` `Node(``1``);``    ``root.left = ``new` `Node(``2``);``    ``root.right = ``new` `Node(``2``);``    ``root.left.left = ``new` `Node(``3``);``    ``root.left.right = ``new` `Node(``4``);``    ``root.right.left = ``new` `Node(``4``);``    ``root.right.right = ``new` `Node(``3``);` `    ``if` `(isSymmetric(root))``      ``System.out.println(``"Symmetric"``);``    ``else``      ``System.out.println(``"Not symmetric"``);``  ``}``}`

## Python3

 `class` `Node:``    ``def` `__init__(``self``, key):``        ``self``.key ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `def` `isSymmetric(root):``    ``# If the root is null, then the binary tree is symmetric``    ``if` `not` `root:``        ``return` `True``    ` `    ``# Create a stack to store the left and right subtrees of the root``    ``stack ``=` `[]``    ``stack.append(root.left)``    ``stack.append(root.right)``    ` `    ``# Continue the loop until the stack is empty``    ``while` `stack:``        ``# Pop the left and right subtrees from the stack``        ``node1 ``=` `stack.pop()``        ``node2 ``=` `stack.pop()``        ` `        ``# If both nodes are null, continue the loop``        ``if` `not` `node1 ``and` `not` `node2:``            ``continue``        ` `        ``# If one of the nodes is null, the binary tree is not symmetric``        ``if` `not` `node1 ``or` `not` `node2:``            ``return` `False``        ` `        ``# If the values of the nodes are not equal, the binary tree is not symmetric``        ``if` `node1.key !``=` `node2.key:``            ``return` `False``        ` `        ``# Push the left and right subtrees of the left and right nodes onto the stack in the opposite order``        ``stack.append(node1.left)``        ``stack.append(node2.right)``        ``stack.append(node1.right)``        ``stack.append(node2.left)``    ` `    ``# If the loop completes, the binary tree is symmetric``    ``return` `True` `# Driver code``root ``=` `Node(``1``)``root.left ``=` `Node(``2``)``root.right ``=` `Node(``2``)``root.left.left ``=` `Node(``3``)``root.left.right ``=` `Node(``4``)``root.right.left ``=` `Node(``4``)``root.right.right ``=` `Node(``3``)` `if` `isSymmetric(root):``    ``print``(``"Symmetric"``)``else``:``    ``print``(``"Not symmetric"``)`

## C#

 `using` `System;``using` `System.Collections;` `public` `class` `Node``{``    ``public` `int` `key;``    ``public` `Node left, right;` `    ``// Constructor``    ``public` `Node(``int` `item)``    ``{``        ``key = item;``        ``left = right = ``null``;``    ``}``}` `public` `class` `GFG``{``    ``// Returns true if a tree is symmetric i.e. mirror image``    ``// of itself``    ``static` `bool` `isSymmetric(Node root)``    ``{``        ``// If the root is null, then the binary tree is``        ``// symmetric.``        ``if` `(root == ``null``)``        ``{``            ``return` `true``;``        ``}` `        ``// Create a stack to store the left and right subtrees``        ``// of the root.``        ``Stack stack = ``new` `Stack();``        ``stack.Push(root.left);``        ``stack.Push(root.right);` `        ``// Continue the loop until the stack is empty.``        ``while` `(stack.Count != 0)``        ``{``            ``// Pop the left and right subtrees from the stack.``            ``Node node1 = (Node)stack.Pop();``            ``Node node2 = (Node)stack.Pop();` `            ``// If both nodes are null, continue the loop.``            ``if` `(node1 == ``null` `&& node2 == ``null``)``            ``{``                ``continue``;``            ``}` `            ``// If one of the nodes is null, the binary tree``            ``// is not symmetric.``            ``if` `(node1 == ``null` `|| node2 == ``null``)``            ``{``                ``return` `false``;``            ``}` `            ``// If the values of the nodes are not equal, the``            ``// binary tree is not symmetric.``            ``if` `(node1.key != node2.key)``            ``{``                ``return` `false``;``            ``}` `            ``// Push the left and right subtrees of the left``            ``// and right nodes onto the stack in the``            ``// opposite order.``            ``stack.Push(node1.left);``            ``stack.Push(node2.right);``            ``stack.Push(node1.right);``            ``stack.Push(node2.left);``        ``}` `        ``// If the loop completes, the binary tree is``        ``// symmetric.``        ``return` `true``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``// Let us construct the Tree shown in the above``        ``// figure``        ``Node root = ``new` `Node(1);``        ``root.left = ``new` `Node(2);``        ``root.right = ``new` `Node(2);``        ``root.left.left = ``new` `Node(3);``        ``root.left.right = ``new` `Node(4);``        ``root.right.left = ``new` `Node(4);``        ``root.right.right = ``new` `Node(3);` `        ``if` `(isSymmetric(root))``            ``Console.WriteLine(``"Symmetric"``);``        ``else``            ``Console.WriteLine(``"Not symmetric"``);``    ``}``}`

## Javascript

 `// Node class for binary tree``class Node {``    ``constructor(item) {``        ``this``.key = item;``        ``this``.left = ``null``;``        ``this``.right = ``null``;``    ``}``}` `// Returns true if a tree is symmetric i.e. mirror image of itself``function` `isSymmetric(root) {``    ``// If the root is null, then the binary tree is symmetric.``    ``if` `(root === ``null``) {``        ``return` `true``;``    ``}` `    ``// Create a stack to store the left and right subtrees of the root.``    ``const stack = [];``    ``stack.push(root.left);``    ``stack.push(root.right);` `    ``// Continue the loop until the stack is empty.``    ``while` `(stack.length > 0) {``        ``// Pop the left and right subtrees from the stack.``        ``const node1 = stack.pop();``        ``const node2 = stack.pop();` `        ``// If both nodes are null, continue the loop.``        ``if` `(node1 === ``null` `&& node2 === ``null``) {``            ``continue``;``        ``}` `        ``// If one of the nodes is null, the binary tree is not symmetric.``        ``if` `(node1 === ``null` `|| node2 === ``null``) {``            ``return` `false``;``        ``}` `        ``// If the values of the nodes are not equal, the binary tree is not symmetric.``        ``if` `(node1.key !== node2.key) {``            ``return` `false``;``        ``}` `        ``// Push the left and right subtrees of the left and right nodes onto the stack in the opposite order.``        ``stack.push(node1.left);``        ``stack.push(node2.right);``        ``stack.push(node1.right);``        ``stack.push(node2.left);``    ``}` `    ``// If the loop completes, the binary tree is symmetric.``    ``return` `true``;``}` `// Driver code``(``function``() {``    ``// Let us construct the Tree shown in the above figure``    ``const root = ``new` `Node(1);``    ``root.left = ``new` `Node(2);``    ``root.right = ``new` `Node(2);``    ``root.left.left = ``new` `Node(3);``    ``root.left.right = ``new` `Node(4);``    ``root.right.left = ``new` `Node(4);``    ``root.right.right = ``new` `Node(3);` `    ``if` `(isSymmetric(root)) {``        ``console.log(``"Symmetric"``);``    ``} ``else` `{``        ``console.log(``"Not symmetric"``);``    ``}``})();`

Output

`Symmetric`

Time Complexity: O(n) where n is the number of nodes.
Space Complexity: O(h) where h is the height of the tree.

Iterative approach using a queue:

`The basic idea is to check if the left and right subtrees of the root node are mirror images of each other. To do this, we perform a level-order traversal of the binary tree using a queue. We push the root node into the queue twice, initially. We dequeue two nodes at a time from the front of the queue and check if they are mirror images of each other.`

Follow the steps to solve the problem:

1. If the root node is NULL, return true as an empty binary tree is considered symmetric.
2. Create a queue and push the root node twice into the queue.
3. While the queue is not empty, dequeue two nodes at a time, one for the left subtree and one for the right subtree.
4. If both the left and right nodes are NULL, continue to the next iteration as the subtrees are considered mirror images of each other.
5. If either the left or right node is NULL, or their data is not equal, return false as they are not mirror images of each other.
6. Push the left and right nodes of the left subtree into the queue, followed by the right and left nodes of the right subtree into the queue.
7. If the queue becomes empty and we have not returned false till now, return true as the binary tree is symmetric.

Below is the implementation of the above approach:

## C++

 `// C++ code to implement the iterative approach using a``// queue``#include ``using` `namespace` `std;` `struct` `Node {``    ``int` `data;``    ``struct` `Node* left;``    ``struct` `Node* right;``    ``Node(``int` `val)``    ``{``        ``data = val;``        ``left = NULL;``        ``right = NULL;``    ``}``};` `bool` `isSymmetric(``struct` `Node* root)``{``    ``if` `(root == NULL) {``        ``return` `true``;``    ``}` `    ``queue q;``    ``q.push(root);``    ``q.push(root);` `    ``while` `(!q.empty()) {``        ``Node* leftNode = q.front();``        ``q.pop();``        ``Node* rightNode = q.front();``        ``q.pop();` `        ``// If both leftNode and rightNode are NULL, continue``        ``// to the next iteration``        ``if` `(leftNode == NULL && rightNode == NULL) {``            ``continue``;``        ``}` `        ``// If either leftNode or rightNode is NULL or their``        ``// data is not equal, return false``        ``if` `(leftNode == NULL || rightNode == NULL``            ``|| leftNode->data != rightNode->data) {``            ``return` `false``;``        ``}` `        ``// Pushing the left and right nodes of the current``        ``// node into the queue``        ``q.push(leftNode->left);``        ``q.push(rightNode->right);``        ``q.push(leftNode->right);``        ``q.push(rightNode->left);``    ``}``    ``return` `true``;``}``// Driver Code``int` `main()``{``    ``// Creating a binary tree``    ``struct` `Node* root = ``new` `Node(5);``    ``root->left = ``new` `Node(1);``    ``root->right = ``new` `Node(1);``    ``root->left->left = ``new` `Node(2);``    ``root->right->right = ``new` `Node(2);` `    ``// Checking if the binary tree is symmetric or not``    ``if` `(isSymmetric(root)) {``        ``cout << ``"The binary tree is symmetric\n"``;``    ``}``    ``else` `{``        ``cout << ``"The binary tree is not symmetric\n"``;``    ``}` `    ``return` `0;``}``// This Code is contributed by Veerendra_Singh_Rajpoot`

Output

```The binary tree is symmetric
```

Time Complexity: O(n),The time complexity of this approach is O(n), where n is the number of nodes in the binary tree.
This is because we visit each node of the binary tree once.

Space Complexity: O(n) , The space complexity of this approach is O(n), where n is the number of nodes in the binary tree.
This is because we store the nodes of the binary tree in a queue.
In the worst-case scenario, the queue can contain all the nodes of the binary tree at the last level, which is approximately n/2 nodes.

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