Given two Binary Trees, returns true if two trees are mirror of each other, else false.
Mirror Tree :
Previously discussed approach is here.
Approach: Find the inorder traversal of both the Binary Trees, and check whether one traversal is reverse of another or not. If they are reverse of each other then the trees are mirror of each other, else not.
Implementation
C++
// C++ code to check two binary trees are// mirror.#include<bits/stdc++.h>using namespace std;struct Node{ int data; Node* left, *right;};// inorder traversal of Binary Treevoid inorder(Node *n, vector<int> &v){ if (n->left != NULL) inorder(n->left, v); v.push_back(n->data); if (n->right != NULL) inorder(n->right, v); }// Checking if binary tree is mirror // of each other or not.bool areMirror(Node* a, Node* b){ if (a == NULL && b == NULL) return true; if (a == NULL || b== NULL) return false; // Storing inorder traversals of both // the trees. vector<int> v1, v2; inorder(a, v1); inorder(b, v2); if (v1.size() != v2.size()) return false; // Comparing the two arrays, if they // are reverse then return 1, else 0 for (int i=0, j=v2.size()-1; j >= 0; i++, j--) if (v1[i] != v2[j]) return false; return true;}// Helper function to allocate a new nodeNode* newNode(int data){ Node* node = new Node; node->data = data; node->left = node->right = NULL; return(node);} // Driver codeint 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 for the above approachimport java.util.*;class GFG { static class Node { public int data; public Node left, right; } // inorder traversal of Binary Tree static void inorder(Node n, List<Integer> v) { if (n.left != null) inorder(n.left, v); v.add(n.data); if (n.right != null) inorder(n.right, v); } // Checking if binary tree is mirror of each other or // not. static boolean areMirror(Node a, Node b) { if (a == null && b == null) return true; if (a == null || b == null) return false; // Storing inorder traversals of both the trees. List<Integer> v1 = new ArrayList<>(); List<Integer> v2 = new ArrayList<>(); inorder(a, v1); inorder(b, v2); if (v1.size() != v2.size()) return false; // Comparing the two arrays, if they are reverse // then return 1, else 0 for (int i = 0, j = v2.size() - 1; j >= 0; i++, j--) { if (v1.get(i) != v2.get(j)) return false; } return true; } // Helper function to allocate a new node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return node; } // Driver code public static void main(String[] args) { 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); if (areMirror(a, b)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by lokeshpotta20. |
Python3
# Python3 code to check two binary trees are# mirror.class Node: def __init__(self, data): self.data = data self.left = None self.right = None # inorder traversal of Binary Treedef inorder(n, v): if (n.left != None): inorder(n.left, v); v.append(n.data); if (n.right != None): inorder(n.right, v); # Checking if binary tree is mirror # of each other or not.def areMirror(a, b): if (a == None and b == None): return True; if (a == None or b== None): return False; # Storing inorder traversals of both # the trees. v1 = [] v2 = [] inorder(a, v1); inorder(b, v2); if (len(v1) != len(v2)): return False; # Comparing the two arrays, if they # are reverse then return 1, else 0 i = 0 j = len(v2) - 1 while j >= 0: if (v1[i] != v2[j]): return False i+=1 j-=1 return True;# Helper function to allocate a new nodedef newNode(data): node = Node(data) return node # Driver codeif __name__=="__main__": a = newNode(1); 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); if areMirror(a, b): print("Yes") else: print("No")# This code is contributed by rutvik_56 |
C#
// C# code to check two binary trees are// mirror.using System;using System.Collections;using System.Collections.Generic;class GFG{ class Node{ public int data; public Node left, right;}; // inorder traversal of Binary Treestatic void inorder(Node n, ref List<int> v){ if (n.left != null) inorder(n.left, ref v); v.Add(n.data); if (n.right != null) inorder(n.right, ref v); } // Checking if binary tree is mirror // of each other or not.static bool areMirror(Node a, Node b){ if (a == null && b == null) return true; if (a == null || b == null) return false; // Storing inorder traversals of both // the trees. List<int> v1 = new List<int>(); List<int> v2 = new List<int>(); inorder(a, ref v1); inorder(b, ref v2); if (v1.Count != v2.Count) return false; // Comparing the two arrays, if they // are reverse then return 1, else 0 for(int i = 0, j = v2.Count - 1; j >= 0; i++, j--) if (v1[i] != v2[j]) return false; return true;} // Helper function to allocate a new nodestatic Node newNode(int data){ Node node = new Node(); node.data = data; node.left = node.right = null; return(node);} // Driver codestatic void Main(string []args){ 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); if (areMirror(a, b)) { Console.Write("Yes"); } else { Console.Write("No"); }}}// This code is contributed by pratham76 |
Javascript
<script>// JavaScript code to check two binary trees are// mirror.class Node{ constructor() { this.data = 0; this.left = null; this.right = null; }}; // inorder traversal of Binary Treefunction inorder(n, v){ if (n.left != null) inorder(n.left, v); v.push(n.data); if (n.right != null) inorder(n.right, v); } // Checking if binary tree is mirror // of each other or not.function areMirror(a, b){ if (a == null && b == null) return true; if (a == null || b == null) return false; // Storing inorder traversals of both // the trees. var v1 = []; var v2 = []; inorder(a, v1); inorder(b, v2); if (v1.length != v2.length) return false; // Comparing the two arrays, if they // are reverse then return 1, else 0 for(var i = 0, j = v2.length - 1; j >= 0; i++, j--) if (v1[i] != v2[j]) return false; return true;} // Helper function to allocate a new nodefunction newNode(data){ var node = new Node(); node.data = data; node.left = node.right = null; return(node);} // Driver codevar a = newNode(1);var 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);if (areMirror(a, b)){ document.write("Yes");}else{ document.write("No");}</script> |
Output
Yes
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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