Given two Binary Trees, write a function that returns true if two trees are mirror of each other, else false. For example, the function should return true for following input trees.
This problem is different from the problem discussed here.
For two trees ‘a’ and ‘b’ to be mirror images, the following three conditions must be true:
- Their root node’s key must be same
- Left subtree of root of ‘a’ and right subtree root of ‘b’ are mirror.
- Right subtree of ‘a’ and left subtree of ‘b’ are mirror.
Below is implementation of above idea.
C++
// C++ program to check if two trees are mirror// of each other#include<bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left child and a pointer to right child */struct Node{ int data; Node* left, *right;};/* Given two trees, return true if they are mirror of each other *//*As function has to return bool value instead integer value*/bool areMirror(Node* a, Node* b){ /* Base case : Both empty */ if (a==NULL && b==NULL) return true; // If only one is empty if (a==NULL || b == NULL) return false; /* Both non-empty, compare them recursively Note that in recursive calls, we pass left of one tree and right of other tree */ return a->data == b->data && areMirror(a->left, b->right) && areMirror(a->right, b->left);}/* Helper function that allocates a new node */Node* newNode(int data){ Node* node = new Node; node->data = data; node->left = node->right = NULL; return(node);}/* Driver program to test areMirror() */int main(){ Node *a = newNode(1); Node *b = newNode(1); a->left = newNode(2); a->right = newNode(3); a->left->left = newNode(4); a->left->right = newNode(5); b->left = newNode(3); b->right = newNode(2); b->right->left = newNode(5); b->right->right = newNode(4); areMirror(a, b)? cout << "Yes" : cout << "No"; return 0;} |
Java
// Java program to see if two trees // are mirror of each other// A binary tree nodeclass Node { int data; Node left, right; public Node(int data) { this.data = data; left = right = null; }}class BinaryTree { Node a, b; /* Given two trees, return true if they are mirror of each other */ boolean areMirror(Node a, Node b) { /* Base case : Both empty */ if (a == null && b == null) return true; // If only one is empty if (a == null || b == null) return false; /* Both non-empty, compare them recursively Note that in recursive calls, we pass left of one tree and right of other tree */ return a.data == b.data && areMirror(a.left, b.right) && areMirror(a.right, b.left); } // Driver code to test above methods public static void main(String[] args) { BinaryTree tree = new BinaryTree(); Node a = new Node(1); Node b = new Node(1); a.left = new Node(2); a.right = new Node(3); a.left.left = new Node(4); a.left.right = new Node(5); b.left = new Node(3); b.right = new Node(2); b.right.left = new Node(5); b.right.right = new Node(4); if (tree.areMirror(a, b) == true) System.out.println("Yes"); else System.out.println("No"); }}// This code has been contributed by Mayank Jaiswal(mayank_24) |
Python3
# Python3 program to check if two# trees are mirror of each other# A binary tree nodeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None# Given two trees, return true # if they are mirror of each otherdef areMirror(a, b): # Base case : Both empty if a is None and b is None: return True # If only one is empty if a is None or b is None: return False # Both non-empty, compare them # recursively. Note that in # recursive calls, we pass left # of one tree and right of other tree return (a.data == b.data and areMirror(a.left, b.right) and areMirror(a.right , b.left))# Driver coderoot1 = Node(1)root2 = Node(1)root1.left = Node(2)root1.right = Node(3)root1.left.left = Node(4)root1.left.right = Node(5)root2.left = Node(3)root2.right = Node(2)root2.right.left = Node(5)root2.right.right = Node(4)if areMirror(root1, root2): print ("Yes")else: print ("No")# This code is contributed by AshishR |
C#
using System;// c# program to see if two trees // are mirror of each other // A binary tree node public class Node{ public int data; public Node left, right; public Node(int data) { this.data = data; left = right = null; }}public class BinaryTree{ public Node a, b; /* Given two trees, return true if they are mirror of each other */ public virtual bool areMirror(Node a, Node b) { /* Base case : Both empty */ if (a == null && b == null) { return true; } // If only one is empty if (a == null || b == null) { return false; } /* Both non-empty, compare them recursively Note that in recursive calls, we pass left of one tree and right of other tree */ return a.data == b.data && areMirror(a.left, b.right) && areMirror(a.right, b.left); } // Driver code to test above methods public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); Node a = new Node(1); Node b = new Node(1); a.left = new Node(2); a.right = new Node(3); a.left.left = new Node(4); a.left.right = new Node(5); b.left = new Node(3); b.right = new Node(2); b.right.left = new Node(5); b.right.right = new Node(4); if (tree.areMirror(a, b) == true) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by Shrikant13 |
Javascript
<script>// javascript program to see if two trees // are mirror of each other// A binary tree nodeclass Node { Node(data) { this.data = data; this.left = this.right = null; }}var a, b; /* * Given two trees, return true if they are mirror of each other */ function areMirror( a, b) { /* Base case : Both empty */ if (a == null && b == null) return true; // If only one is empty if (a == null || b == null) return false; /* * Both non-empty, compare them recursively Note that in recursive calls, we * pass left of one tree and right of other tree */ return a.data == b.data && areMirror(a.left, b.right) && areMirror(a.right, b.left); } // Driver code to test above methods a = new Node(1); b = new Node(1); left = new Node(2); right = new Node(3); left.left = new Node(4); left.right = new Node(5); left = new Node(3); right = new Node(2); right.left = new Node(5); right.right = new Node(4); if (areMirror(a, b) == true) document.write("Yes"); else document.write("No");// This code contributed by umadevi9616 </script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(h) where h is height of binary tree
Approach 2: Iterative Solution
Another approach to check if two trees are mirrors of each other is to use an iterative algorithm that uses a stack to keep track of the nodes being traversed. This algorithm is similar to the iterative algorithm for checking if a tree is symmetric, but with a slight modification to check for mirror symmetry.
This solution uses two stacks to iterate over the two trees in a synchronized way. It pops a node from both stacks, checks if their data is equal, and then pushes their left and right children onto the stacks in the opposite order. The algorithm terminates early if at any point the nodes being compared have unequal data or unequal numbers of children.
Here is the implementation for the iterative solution:
C++
#include <bits/stdc++.h>using namespace std;// A binary tree nodeclass Node {public: int data; Node* left; Node* right; Node(int d) { this->data = d; this->left = NULL; this->right = NULL; }};bool isMirrorIterative(Node* root1, Node* root2){ if (root1 == NULL && root2 == NULL) return true; if (root1 == NULL || root2 == NULL) return false; stack<Node*> s1, s2; s1.push(root1); s2.push(root2); while (!s1.empty() && !s2.empty()) { Node* curr1 = s1.top(); s1.pop(); Node* curr2 = s2.top(); s2.pop(); if (curr1->data != curr2->data) return false; if (curr1->left != NULL && curr2->right != NULL) { s1.push(curr1->left); s2.push(curr2->right); } else if (curr1->left != NULL || curr2->right != NULL) return false; if (curr1->right != NULL && curr2->left != NULL) { s1.push(curr1->right); s2.push(curr2->left); } else if (curr1->right != NULL || curr2->left != NULL) return false; } if (!s1.empty() || !s2.empty()) return false; return true;}int main(){ // create the two trees Node* root1 = new Node(1); root1->left = new Node(2); root1->right = new Node(3); root1->left->left = new Node(4); root1->left->right = new Node(5); Node* root2 = new Node(1); root2->left = new Node(3); root2->right = new Node(2); root2->right->left = new Node(5); root2->right->right = new Node(4); // check if the two trees are mirrors if (isMirrorIterative(root1, root2)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
import java.util.*;// A binary tree nodeclass Node {int data;Node left;Node right; Node(int d) { this.data = d; this.left = null; this.right = null;}}class Main {static boolean isMirrorIterative(Node root1, Node root2) {if (root1 == null && root2 == null)return true; if (root1 == null || root2 == null) return false; Stack<Node> s1 = new Stack<>(); Stack<Node> s2 = new Stack<>(); s1.push(root1); s2.push(root2); while (!s1.empty() && !s2.empty()) { Node curr1 = s1.pop(); Node curr2 = s2.pop(); if (curr1.data != curr2.data) return false; if (curr1.left != null && curr2.right != null) { s1.push(curr1.left); s2.push(curr2.right); } else if (curr1.left != null || curr2.right != null) return false; if (curr1.right != null && curr2.left != null) { s1.push(curr1.right); s2.push(curr2.left); } else if (curr1.right != null || curr2.left != null) return false; } if (!s1.empty() || !s2.empty()) return false; return true;}public static void main(String[] args) { // create the two trees Node root1 = new Node(1); root1.left = new Node(2); root1.right = new Node(3); root1.left.left = new Node(4); root1.left.right = new Node(5); Node root2 = new Node(1); root2.left = new Node(3); root2.right = new Node(2); root2.right.left = new Node(5); root2.right.right = new Node(4); // check if the two trees are mirrors if (isMirrorIterative(root1, root2)) System.out.println("Yes"); else System.out.println("No");}} |
Python3
# Python program for the above approachclass Node: def __init__(self, d): self.data = d self.left = None self.right = None# Function to check if two binary trees are mirrors of each otherdef isMirrorIterative(root1, root2): # If both trees are empty, they are mirrors if root1 is None and root2 is None: return True # If one tree is empty and the other is not, they are not mirrors if root1 is None or root2 is None: return False stack1 = [] stack2 = [] stack1.append(root1) stack2.append(root2) while len(stack1) > 0 and len(stack2) > 0: curr1 = stack1.pop() curr2 = stack2.pop() # If data of current nodes doesn't match, they are not mirrors if curr1.data != curr2.data: return False # Check the left child of first tree and the right child of the second tree if curr1.left is not None and curr2.right is not None: stack1.append(curr1.left) stack2.append(curr2.right) elif curr1.left is not None or curr2.right is not None: return False # Check the right child of first tree and the left child of the second tree if curr1.right is not None and curr2.left is not None: stack1.append(curr1.right) stack2.append(curr2.left) elif curr1.right is not None or curr2.left is not None: return False # If any stack is not empty, trees are not mirrors if len(stack1) > 0 or len(stack2) > 0: return False # Trees are mirrors return True# create the two treesroot1 = Node(1)root1.left = Node(2)root1.right = Node(3)root1.left.left = Node(4)root1.left.right = Node(5)root2 = Node(1)root2.left = Node(3)root2.right = Node(2)root2.right.left = Node(5)root2.right.right = Node(4)# check if the two trees are mirrorsif isMirrorIterative(root1, root2): print("Yes") # Output: Yeselse: print("No") # Output: No |
Javascript
// JavaScript Program for the above approachclass Node { constructor(d) { this.data = d; this.left = null; this.right = null; }}// Function to check if two binary trees are mirrors of each otherfunction isMirrorIterative(root1, root2) { // If both trees are empty, they are mirrors if (root1 === null && root2 === null) return true; // If one tree is empty and the other is not, they are not mirrors if (root1 === null || root2 === null) return false; const stack1 = []; const stack2 = []; stack1.push(root1); stack2.push(root2); while (stack1.length > 0 && stack2.length > 0) { const curr1 = stack1.pop(); const curr2 = stack2.pop(); // If data of current nodes doesn't match, they are not mirrors if (curr1.data !== curr2.data) return false; // Check the left child of first tree and the right child of the second tree if (curr1.left !== null && curr2.right !== null) { stack1.push(curr1.left); stack2.push(curr2.right); } else if (curr1.left !== null || curr2.right !== null) return false; // Check the right child of first tree and the left child of the second tree if (curr1.right !== null && curr2.left !== null) { stack1.push(curr1.right); stack2.push(curr2.left); } else if (curr1.right !== null || curr2.left !== null) return false; } // If any stack is not empty, trees are not mirrors if (stack1.length > 0 || stack2.length > 0) return false; // Trees are mirrors return true;}// create the two treesconst root1 = new Node(1);root1.left = new Node(2);root1.right = new Node(3);root1.left.left = new Node(4);root1.left.right = new Node(5);const root2 = new Node(1);root2.left = new Node(3);root2.right = new Node(2);root2.right.left = new Node(5);root2.right.right = new Node(4);// check if the two trees are mirrorsif (isMirrorIterative(root1, root2)) console.log("Yes"); // Output: Yeselse console.log("No"); // Output: No// THIS CODE IS CONTRIBUTED BY PIYUSH AGARWAL |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(h) where h is height of binary tree
Iterative method to check if two trees are mirror of each other
This article is contributed by Ashish Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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