Check if two trees are Mirror

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: 

1. Their root node’s key must be same
2. Left subtree of root of ‘a’ and right subtree root of ‘b’ are mirror.
3. 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 node
class 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 node
class 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 other
def 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 code
root1 = 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 node
class 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 C++ code for the iterative solution:

C++

#include <bits/stdc++.h>
using namespace std;

// A binary tree node
class 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 node
class 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");
}
}

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.