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Check if two trees are mirror of each other using level order traversal

  • Difficulty Level : Medium
  • Last Updated : 18 Aug, 2021

Given two binary trees, the task is to check whether the two binary trees is a mirror of each other or not. 
Mirror of a Binary Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged.
 

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Trees in the above figure are mirrors of each other.
 



A recursive solution and an iterative method using inorder traversal to check whether the two binary trees is a mirror of each other or not have been already discussed. In this post a solution using level order traversal has been discussed.
The idea is to use a queue in which two nodes of both the trees which needs to be checked for equality are present together. At each step of level order traversal, get two nodes from the queue, check for their equality and then insert next two children nodes of these nodes which need to be checked for equality. During insertion step, first left child of first tree node and right child of second tree node are inserted. After this right child of first tree node and left child of second tree node are inserted. If at any stage one node is NULL and other is not, then both trees are not a mirror of each other.
Below is the implementation of above approach:
 

C++




// C++ implementation to check whether the two
// binary trees are mirrors of each other or not
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a node in binary tree
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Function to create and return
// a new node for a binary tree
struct Node* newNode(int data)
{
    struct Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to check whether the two binary trees
// are mirrors of each other or not
string areMirrors(Node* a, Node* b)
{
    // If both are NULL, then are mirror.
    if (a == NULL && b == NULL)
        return "Yes";
 
    // If only one is NULL, then not
    // mirror.
    if (a == NULL || b == NULL)
        return "No";
 
    queue<Node*> q;
 
    // Push root of both trees in queue.
    q.push(a);
    q.push(b);
 
    while (!q.empty()) {
 
        // Pop two elements of queue, to
        // get two nodes and check if they
        // are symmetric.
        a = q.front();
        q.pop();
 
        b = q.front();
        q.pop();
 
        // If data value of both nodes is
        // not same, then not mirror.
        if (a->data != b->data)
            return "No";
 
        // Push left child of first tree node
        // and right child of second tree node
        // into queue if both are not NULL.
        if (a->left && b->right) {
            q.push(a->left);
            q.push(b->right);
        }
 
        // If any one of the nodes is NULL and
        // other is not NULL, then not mirror.
        else if (a->left || b->right)
            return "No";
 
        // Push right child of first tree node
        // and left child of second tree node
        // into queue if both are not NULL.
        if (a->right && b->left) {
            q.push(a->right);
            q.push(b->left);
        }
 
        // If any one of the nodes is NULL and
        // other is not NULL, then not mirror.
        else if (a->right || b->left)
            return "No";
    }
 
    return "Yes";
}
// Driver Code
int main()
{
    // 1st binary tree formation
    /*
            1
           / \
          3   2
             / \
            5   4
        */
    Node* root1 = newNode(1);
    root1->left = newNode(3);
    root1->right = newNode(2);
    root1->right->left = newNode(5);
    root1->right->right = newNode(4);
 
    // 2nd binary tree formation
    /*
            1
           / \
          2   3
         / \
        4   5
        */
    Node* root2 = newNode(1);
    root2->left = newNode(2);
    root2->right = newNode(3);
    root2->left->left = newNode(4);
    root2->left->right = newNode(5);
 
    cout << areMirrors(root1, root2);
    return 0;
}

Java




// Java implementation to check whether the two
// binary trees are mirrors of each other or not
import java.util.*;
 
class GFG
{
 
// Structure of a node in binary tree
static class Node
{
    int data;
    Node left, right;
};
 
// Function to create and return
// a new node for a binary tree
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to check whether the two binary trees
// are mirrors of each other or not
static String areMirrors(Node a, Node b)
{
    // If both are null, then are mirror.
    if (a == null && b == null)
        return "Yes";
 
    // If only one is null, then not
    // mirror.
    if (a == null || b == null)
        return "No";
 
    Queue<Node> q = new LinkedList<Node>();
 
    // Push root of both trees in queue.
    q.add(a);
    q.add(b);
 
    while (q.size() > 0)
    {
 
        // remove two elements of queue, to
        // get two nodes and check if they
        // are symmetric.
        a = q.peek();
        q.remove();
 
        b = q.peek();
        q.remove();
 
        // If data value of both nodes is
        // not same, then not mirror.
        if (a.data != b.data)
            return "No";
 
        // Push left child of first tree node
        // and right child of second tree node
        // into queue if both are not null.
        if (a.left != null && b.right != null)
        {
            q.add(a.left);
            q.add(b.right);
        }
 
        // If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.left != null || b.right != null)
            return "No";
 
        // Push right child of first tree node
        // and left child of second tree node
        // into queue if both are not null.
        if (a.right != null && b.left != null)
        {
            q.add(a.right);
            q.add(b.left);
        }
 
        //If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.right != null || b.left != null)
            return "No";
    }
 
    return "Yes";
}
 
// Driver Code
public static void main(String args[])
{
    // 1st binary tree formation
    /*
            1
        / \
        3 2
            / \
            5 4
        */
    Node root1 = newNode(1);
    root1.left = newNode(3);
    root1.right = newNode(2);
    root1.right.left = newNode(5);
    root1.right.right = newNode(4);
 
    // 2nd binary tree formation
    /*
            1
        / \
        2 3
        / \
        4 5
        */
    Node root2 = newNode(1);
    root2.left = newNode(2);
    root2.right = newNode(3);
    root2.left.left = newNode(4);
    root2.left.right = newNode(5);
 
    System.out.print(areMirrors(root1, root2));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 implementation to check whether the two
# binary trees are mirrors of each other or not
 
# Structure of a node in binary tree
class Node: 
     
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
     
# Function to check whether the two binary
# trees are mirrors of each other or not
def areMirrors(a, b):
  
    # If both are NULL, then are mirror.
    if a == None and b == None:
        return "Yes"
 
    # If only one is NULL, then not mirror.
    if a == None or b == None:
        return "No"
 
    q = []
 
    # append root of both trees in queue.
    q.append(a)
    q.append(b)
 
    while len(q) > 0
 
        # Pop two elements of queue,
        # to get two nodes and check
        # if they are symmetric.
        a = q.pop(0)
        b = q.pop(0)
 
        # If data value of both nodes is
        # not same, then not mirror.
        if a.data != b.data:
            return "No"
 
        # append left child of first tree node
        # and right child of second tree node
        # into queue if both are not NULL.
        if a.left and b.right: 
            q.append(a.left)
            q.append(b.right)
 
        # If any one of the nodes is NULL and
        # other is not NULL, then not mirror.
        elif a.left or b.right:
            return "No"
 
        # Append right child of first tree node
        # and left child of second tree node
        # into queue if both are not NULL.
        if a.right and b.left: 
            q.append(a.right)
            q.append(b.left)
 
        # If any one of the nodes is NULL and
        # other is not NULL, then not mirror.
        elif a.right or b.left:
            return "No"
      
    return "Yes"
  
# Driver Code
if __name__ == "__main__":
  
    # 1st binary tree formation
    root1 = Node(1)
    root1.left = Node(3)
    root1.right = Node(2)
    root1.right.left = Node(5)
    root1.right.right = Node(4)
 
    # 2nd binary tree formation
    root2 = Node(1)
    root2.left = Node(2)
    root2.right = Node(3)
    root2.left.left = Node(4)
    root2.left.right = Node(5)
 
    print(areMirrors(root1, root2))
 
# This code is contributed by Rituraj Jain

C#




// C# implementation to check whether the two
// binary trees are mirrors of each other or not
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Structure of a node in binary tree
public class Node
{
    public int data;
    public Node left, right;
};
 
// Function to create and return
// a new node for a binary tree
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to check whether the two binary trees
// are mirrors of each other or not
static String areMirrors(Node a, Node b)
{
    // If both are null, then are mirror.
    if (a == null && b == null)
        return "Yes";
 
    // If only one is null, then not
    // mirror.
    if (a == null || b == null)
        return "No";
 
    Queue<Node> q = new Queue<Node>();
 
    // Push root of both trees in queue.
    q.Enqueue(a);
    q.Enqueue(b);
 
    while (q.Count > 0)
    {
 
        // remove two elements of queue, to
        // get two nodes and check if they
        // are symmetric.
        a = q.Peek();
        q.Dequeue();
 
        b = q.Peek();
        q.Dequeue();
 
        // If data value of both nodes is
        // not same, then not mirror.
        if (a.data != b.data)
            return "No";
 
        // Push left child of first tree node
        // and right child of second tree node
        // into queue if both are not null.
        if (a.left != null && b.right != null)
        {
            q.Enqueue(a.left);
            q.Enqueue(b.right);
        }
 
        // If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.left != null || b.right != null)
            return "No";
 
        // Push right child of first tree node
        // and left child of second tree node
        // into queue if both are not null.
        if (a.right != null && b.left != null)
        {
            q.Enqueue(a.right);
            q.Enqueue(b.left);
        }
 
        //If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.right != null || b.left != null)
            return "No";
    }
    return "Yes";
}
 
// Driver Code
public static void Main(String []args)
{
    // 1st binary tree formation
    /*
            1
        / \
        3 2
            / \
            5 4
        */
    Node root1 = newNode(1);
    root1.left = newNode(3);
    root1.right = newNode(2);
    root1.right.left = newNode(5);
    root1.right.right = newNode(4);
 
    // 2nd binary tree formation
    /*
            1
        / \
        2 3
        / \
        4 5
        */
    Node root2 = newNode(1);
    root2.left = newNode(2);
    root2.right = newNode(3);
    root2.left.left = newNode(4);
    root2.left.right = newNode(5);
 
    Console.Write(areMirrors(root1, root2));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// JavaScript implementation to check whether the two
// binary trees are mirrors of each other or not
 
// Structure of a node in binary tree
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};
 
// Function to create and return
// a new node for a binary tree
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to check whether the two binary trees
// are mirrors of each other or not
function areMirrors(a, b)
{
    // If both are null, then are mirror.
    if (a == null && b == null)
        return "Yes";
 
    // If only one is null, then not
    // mirror.
    if (a == null || b == null)
        return "No";
 
    var q = [];
 
    // Push root of both trees in queue.
    q.push(a);
    q.push(b);
 
    while (q.length > 0)
    {
 
        // remove two elements of queue, to
        // get two nodes and check if they
        // are symmetric.
        a = q[0];
        q.shift();
 
        b = q[0];
        q.shift();
 
        // If data value of both nodes is
        // not same, then not mirror.
        if (a.data != b.data)
            return "No";
 
        // Push left child of first tree node
        // and right child of second tree node
        // into queue if both are not null.
        if (a.left != null && b.right != null)
        {
            q.push(a.left);
            q.push(b.right);
        }
 
        // If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.left != null || b.right != null)
            return "No";
 
        // Push right child of first tree node
        // and left child of second tree node
        // into queue if both are not null.
        if (a.right != null && b.left != null)
        {
            q.push(a.right);
            q.push(b.left);
        }
 
        //If any one of the nodes is null and
        // other is not null, then not mirror.
        else if (a.right != null || b.left != null)
            return "No";
    }
    return "Yes";
}
 
// Driver Code
// 1st binary tree formation
/*
        1
    / \
    3 2
        / \
        5 4
    */
var root1 = newNode(1);
root1.left = newNode(3);
root1.right = newNode(2);
root1.right.left = newNode(5);
root1.right.right = newNode(4);
// 2nd binary tree formation
/*
        1
    / \
    2 3
    / \
    4 5
    */
var root2 = newNode(1);
root2.left = newNode(2);
root2.right = newNode(3);
root2.left.left = newNode(4);
root2.left.right = newNode(5);
document.write(areMirrors(root1, root2));
 
 
</script>
Output: 
Yes

 

Time complexity: O(N)
Auxiliary Space: O(N)  




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