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 Nodestruct Node { int key; struct Node *left, *right;};// Utility function to create new NodeNode* 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// mirrorbool 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 itselfbool isSymmetric(struct Node* root){ // Check if tree is mirror of itself return isMirror(root, root);}// Driver codeint 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 Nodetypedef struct Node { int key; struct Node *left, *right;} Node;// Utility function to create new NodeNode* 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// mirrorbool 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// itselfbool isSymmetric(Node* root){ // Check if tree is mirror of itself return isMirror(root, root);}// Driver codeint 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 notclass 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 structureclass 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 mirrordef 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 Falsedef isSymmetric(root): # Check if tree is mirror of itself return isMirror(root, root)# Driver Code# Let's construct the tree show in the above figureroot = 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 notusing 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 notclass 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 mirrorfunction 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 itselffunction isSymmetric(){ // check if tree is mirror of itself return isMirror(root, root);}// Driver Coderoot = 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> |
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:
- Create a stack and push the root node onto it twice.
- 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. - 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 <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { int key; struct Node *left, *right;};// Utility function to create new NodeNode* 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// itselfbool 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<Node*> 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 codeint 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 notimport java.util.*;// A Binary Tree Nodeclass 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<Node> 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 = Nonedef 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 coderoot = 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 treeclass Node { constructor(item) { this.key = item; this.left = null; this.right = null; }}// Returns true if a tree is symmetric i.e. mirror image of itselffunction 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:
- If the root node is NULL, return true as an empty binary tree is considered symmetric.
- Create a queue and push the root node twice into the queue.
- While the queue is not empty, dequeue two nodes at a time, one for the left subtree and one for the right subtree.
- If both the left and right nodes are NULL, continue to the next iteration as the subtrees are considered mirror images of each other.
- 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.
- 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.
- 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 <bits/stdc++.h>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<Node*> 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 Codeint 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 |
Java
//Java code to implement the iterative approach using a queueimport java.util.LinkedList;import java.util.Queue;class Node { int data; Node left; Node right; Node(int val) { data = val; left = null; right = null; }}public class SymmetricBinaryTree { //Function to implement the iterative approach using a queue public static boolean isSymmetric(Node root) { if (root == null) { return true; } Queue<Node> q = new LinkedList<>(); q.add(root); q.add(root); while (!q.isEmpty()) { Node leftNode = q.poll(); Node rightNode = q.poll(); // 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.add(leftNode.left); q.add(rightNode.right); q.add(leftNode.right); q.add(rightNode.left); } return true; }//Driver Code public static void main(String[] args) { // Creating a binary tree 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)) { System.out.println("The binary tree is symmetric"); } else { System.out.println("The binary tree is not symmetric"); } }}//This code is contributed by Veerendra_Singh_Rajpoot |
Python3
from queue import Queueclass Node: def __init__(self, val): self.data = val self.left = None self.right = Nonedef isSymmetric(root): if root is None: return True q = Queue() q.put(root) q.put(root) while not q.empty(): leftNode = q.get() rightNode = q.get() # If both leftNode and rightNode are None, continue # to the next iteration if leftNode is None and rightNode is None: continue # If either leftNode or rightNode is None or their # data is not equal, return False if leftNode is None or rightNode is None or leftNode.data != rightNode.data: return False # Pushing the left and right nodes of the current # node into the queue q.put(leftNode.left) q.put(rightNode.right) q.put(leftNode.right) q.put(rightNode.left) return Trueif __name__ == '__main__': # Creating a binary tree root = Node(5) root.left = Node(1) root.right = Node(1) root.left.left = Node(2) root.right.right = Node(2) # Checking if the binary tree is symmetric or not if isSymmetric(root): print("The binary tree is symmetric") else: print("The binary tree is not symmetric") |
C#
using System;using System.Collections.Generic;public class Node{ public int Data; public Node Left; public Node Right; public Node(int val) { Data = val; Left = null; Right = null; }}public class BinaryTree{ public static bool IsSymmetric(Node root) { if (root == null) { return true; } Queue<Node> queue = new Queue<Node>(); queue.Enqueue(root); queue.Enqueue(root); while (queue.Count > 0) { Node leftNode = queue.Dequeue(); Node rightNode = queue.Dequeue(); // 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 queue.Enqueue(leftNode.Left); queue.Enqueue(rightNode.Right); queue.Enqueue(leftNode.Right); queue.Enqueue(rightNode.Left); } return true; } public static void Main(string[] args) { // Creating a binary tree 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)) { Console.WriteLine("The binary tree is symmetric"); } else { Console.WriteLine("The binary tree is not symmetric"); } }} |
Javascript
//Javascript code for the above approachclass Node { constructor(val) { this.data = val; this.left = null; this.right = null; }}function isSymmetric(root) { if (root === null) { return true; } const q = []; q.push(root); q.push(root); while (q.length > 0) { const leftNode = q.shift(); const rightNode = q.shift(); // 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 Codefunction main() { // Creating a binary tree const 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)) { console.log("The binary tree is symmetric"); } else { console.log("The binary tree is not symmetric"); }}main(); |
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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