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
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++14
// C++ program to check if a given Binary Tree is symmetric // or not #include <bits/stdc++.h> 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 |
C
// C program to check if a given Binary Tree is symmetric // or not #include <stdbool.h> #include <stdio.h> #include <stdlib.h> // 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 |
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) |
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 |
Javascript
<script> // Javascript program to check is binary // tree is symmetric or not class Node { constructor(item) { this .key = item; this .left = null ; this .right = null ; } } var root = null ; // returns true if trees with roots // as root1 and root2 are mirror function isMirror(node1, 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 function isSymmetric() { // check if tree is mirror of itself return isMirror(root, root); } // Driver Code 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); var output = isSymmetric(); if (output == true ) document.write( "Symmetric" ); else document.write( "Not symmetric" ); // This code is contributed by rrrtnx. </script> |
Symmetric
Time Complexity: O(N)
Auxiliary Space: O(h) where h is the maximum height of the tree
Check for Symmetric Binary Tree (Iterative Approach)
This article is contributed by Muneer Ahmed. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above