Check if two trees are Mirror | Set 2

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Given two Binary Trees, returns true if two trees are mirror of each other, else false.

Mirror Tree :

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 Tree
void 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 node
Node* newNode(int data)
{
  Node* node = new Node;
  node->data  = data;
  node->left  =  node->right  = NULL;
   
  return(node);
}
  
// Driver code
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 for the above approach
import 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 Tree
def 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 node
def newNode(data):
    node = Node(data)
    return node
      
# Driver code
if __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 Tree
static 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 node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}
   
// Driver code
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))
    {
        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 Tree
function 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 node
function newNode(data)
{
    var node = new Node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}
   
// Driver code
var 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)

This article is contributed by Anuj Chauhan. If you like zambiatek and would like to contribute, you can also write an article using write.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek main page and help other Geeks.

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