# 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: Please solve it on “PRACTICE” first, before moving on to the solution.

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 mirror. 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 ` `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 emptu, 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 neither 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 program ` `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); ` ` `  `    ``cout << isSymmetric(root); ` `    ``return` `0; ` `} `

## 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 neither 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(Node node)  ` `    ``{ ` `        ``// check if tree is mirror of itself ` `        ``return` `isMirror(root, root); ` `    ``} ` `  `  `    ``// Driver program ` `    ``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(tree.root); ` `        ``if` `(output == ``true``) ` `            ``System.out.println(``"1"``); ` `        ``else` `            ``System.out.println(``"0"``); ` `    ``} ` `} ` `  `  `// This code has been contributed by Mayank Jaiswal `

## Python

 `# 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 neither 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 Program  ` `# 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` `"1"` `if` `isSymmetric(root) ``=``=` `True` `else` `"0"`  ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## 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 neither 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(Node node)  ` `    ``{  ` `        ``// 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(tree.root);  ` `        ``if` `(output == ``true``)  ` `            ``Console.WriteLine(``"1"``);  ` `        ``else` `            ``Console.WriteLine(``"0"``);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

Output:

`1`

Check for Symmetric Binary Tree (Iterative Approach)

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Improved By : andrew1234, ShubhamDubey7

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