Iterative function to check if two trees are identical

Two trees are identical when they have same data and arrangement of data is also same. To identify if two trees are identical, we need to traverse both trees simultaneously, and while traversing we need to compare data and children of the trees. 
Examples:

Input : Roots of below trees
    10           10
  /   \         /
 5     6       5 
Output : false

Input : Roots of below trees
    10            10
  /   \         /   \
 5     6       5     6
Output : true

We have discussed recursive solution here. In this article iterative solution is discussed. 
The idea is to use level order traversal. We traverse both trees simultaneously and compare the data whenever we dequeue and item from queue. Below is the implementation of the idea. 

C++

/* Iterative C++ program to check if two */
#include <bits/stdc++.h>
using namespace std;

// A Binary Tree Node
struct Node
{
    int data;
    struct Node *left, *right;
};

// Iterative method to find height of Binary Tree
bool areIdentical(Node *root1, Node *root2)
{
    // Return true if both trees are empty
    if (root1==NULL  && root2==NULL) return true;

    // Return false if one is empty and other is not
    if (root1 == NULL) return false;
    if (root2 == NULL) return false;

    // Create an empty queues for simultaneous traversals
    queue<Node *> q1, q2;

    // Enqueue Roots of trees in respective queues
    q1.push(root1);
    q2.push(root2);

    while (!q1.empty() && !q2.empty())
    {
        // Get front nodes and compare them
        Node *n1 = q1.front();
        Node *n2 = q2.front();

        if (n1->data != n2->data)
           return false;

        // Remove front nodes from queues
        q1.pop(), q2.pop();

        /* Enqueue left children of both nodes */
        if (n1->left && n2->left)
        {
            q1.push(n1->left);
            q2.push(n2->left);
        }

        // If one left child is empty and other is not
        else if (n1->left || n2->left)
            return false;

        // Right child code (Similar to left child code)
        if (n1->right && n2->right)
        {
            q1.push(n1->right);
            q2.push(n2->right);
        }
        else if (n1->right || n2->right)
            return false;
    }

    return true;
}

// Utility function to create a new tree node
Node* newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// Driver program to test above functions
int main()
{
    Node *root1 = newNode(1);
    root1->left = newNode(2);
    root1->right = newNode(3);
    root1->left->left = newNode(4);
    root1->left->right = newNode(5);

    Node *root2 = newNode(1);
    root2->left = newNode(2);
    root2->right = newNode(3);
    root2->left->left = newNode(4);
    root2->left->right = newNode(5);

    areIdentical(root1, root2)? cout << "Yes"
                              : cout << "No";
    return 0;
}

Java

/* Iterative Java program to check if two */
import java.util.*;
class GfG {

// A Binary Tree Node
static class Node
{
    int data;
    Node left, right;
}

// Iterative method to find height of Binary Tree
static boolean areIdentical(Node root1, Node root2)
{
    // Return true if both trees are empty
    if (root1 == null && root2 == null)  return true;

    // Return false if one is empty and other is not
    if (root1 == null || root2 == null) return false;

    // Create an empty queues for simultaneous traversals
    Queue<Node > q1 = new LinkedList<Node> ();
    Queue<Node>  q2 = new LinkedList<Node> ();

    // Enqueue Roots of trees in respective queues
    q1.add(root1);
    q2.add(root2);

    while (!q1.isEmpty() && !q2.isEmpty())
    {
        // Get front nodes and compare them
        Node n1 = q1.peek();
        Node n2 = q2.peek();

        if (n1.data != n2.data)
        return false;

        // Remove front nodes from queues
        q1.remove();
        q2.remove();

        /* Enqueue left children of both nodes */
        if (n1.left != null && n2.left != null)
        {
            q1.add(n1.left);
            q2.add(n2.left);
        }

        // If one left child is empty and other is not
        else if (n1.left != null || n2.left != null)
            return false;

        // Right child code (Similar to left child code)
        if (n1.right != null && n2.right != null)
        {
            q1.add(n1.right);
            q2.add(n2.right);
        }
        else if (n1.right != null || n2.right != null)
            return false;
    }

    return true;
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver program to test above functions
public static void main(String[] args)
{
    Node root1 = newNode(1);
    root1.left = newNode(2);
    root1.right = newNode(3);
    root1.left.left = newNode(4);
    root1.left.right = newNode(5);

    Node root2 = newNode(1);
    root2.left = newNode(2);
    root2.right = newNode(3);
    root2.left.left = newNode(4);
    root2.left.right = newNode(5);

    if(areIdentical(root1, root2) == true)
    System.out.println("Yes");
    else
    System.out.println("No");
}
}

Python3

# Iterative Python3 program to check
# if two trees are identical
from queue import Queue

# Utility function to create a
# new tree node
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

# Iterative method to find height of
# Binary Tree
def areIdentical(root1, root2):

    # Return true if both trees are empty
    if (root1 and root2):
        return True

    # Return false if one is empty and
    # other is not
    if (root1 or root2):
        return False

    # Create an empty queues for
    # simultaneous traversals
    q1 = Queue()
    q2 = Queue()

    # Enqueue Roots of trees in
    # respective queues
    q1.put(root1)
    q2.put(root2)

    while (not q1.empty() and not q2.empty()):

        # Get front nodes and compare them
        n1 = q1.queue[0]
        n2 = q2.queue[0]

        if (n1.data != n2.data):
            return False

        # Remove front nodes from queues
        q1.get()
        q2.get()

        # Enqueue left children of both nodes
        if (n1.left and n2.left):
            q1.put(n1.left)
            q2.put(n2.left)

        # If one left child is empty and
        # other is not
        elif (n1.left or n2.left):
            return False

        # Right child code (Similar to
        # left child code)
        if (n1.right and n2.right):
            q1.put(n1.right)
            q2.put(n2.right)
        elif (n1.right or n2.right):
            return False

    return True

# Driver Code
if __name__ == '__main__':
    root1 = newNode(1)
    root1.left = newNode(2)
    root1.right = newNode(3)
    root1.left.left = newNode(4)
    root1.left.right = newNode(5)

    root2 = newNode(1)
    root2.left = newNode(2)
    root2.right = newNode(3)
    root2.left.left = newNode(4)
    root2.left.right = newNode(5)

    if areIdentical(root1, root2):
        print("Yes")
    else:
        print("No")

# This code is contributed by PranchalK

C#

/* Iterative C# program to check if two */
using System;
using System.Collections.Generic;

class GfG
{

// A Binary Tree Node
class Node
{
    public int data;
    public Node left, right;
}

// Iterative method to find height of Binary Tree
static bool areIdentical(Node root1, Node root2)
{
    // Return true if both trees are empty
    if (root1 == null && root2 == null)
        return true;

    // Return false if one is empty and other is not
    if (root1 == null || root2 == null)
        return false;

    // Create an empty queues for
    // simultaneous traversals
    Queue<Node> q1 = new Queue<Node> ();
    Queue<Node> q2 = new Queue<Node> ();

    // Enqueue Roots of trees in respective queues
    q1.Enqueue(root1);
    q2.Enqueue(root2);

    while (q1.Count != 0 && q2.Count != 0)
    {
        // Get front nodes and compare them
        Node n1 = q1.Peek();
        Node n2 = q2.Peek();

        if (n1.data != n2.data)
        return false;

        // Remove front nodes from queues
        q1.Dequeue();
        q2.Dequeue();

        /* Enqueue left children of both nodes */
        if (n1.left != null && n2.left != null)
        {
            q1.Enqueue(n1.left);
            q2.Enqueue(n2.left);
        }

        // If one left child is empty and other is not
        else if (n1.left != null || n2.left != null)
            return false;

        // Right child code (Similar to left child code)
        if (n1.right != null && n2.right != null)
        {
            q1.Enqueue(n1.right);
            q2.Enqueue(n2.right);
        }
        else if (n1.right != null || n2.right != null)
            return false;
    }

    return true;
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver code
public static void Main(String[] args)
{
    Node root1 = newNode(1);
    root1.left = newNode(2);
    root1.right = newNode(3);
    root1.left.left = newNode(4);
    root1.left.right = newNode(5);

    Node root2 = newNode(1);
    root2.left = newNode(2);
    root2.right = newNode(3);
    root2.left.left = newNode(4);
    root2.left.right = newNode(5);

    if(areIdentical(root1, root2) == true)
    Console.WriteLine("Yes");
    else
    Console.WriteLine("No");
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

/* Iterative Javascript program to check if two */

// A Binary Tree Node
class Node
{
  constructor()
  {
    this.data = 0;
    this.left = null;
    this.right = null;
  }
}

// Iterative method to find height of Binary Tree
function areIdentical(root1, root2)
{
    // Return true if both trees are empty
    if (root1 == null && root2 == null)
        return true;

    // Return false if one is empty and other is not
    if (root1 == null || root2 == null)
        return false;

    // Create an empty queues for
    // simultaneous traversals
    var q1 = [];
    var q2 = [];

    // push Roots of trees in respective queues
    q1.push(root1);
    q2.push(root2);

    while (q1.length != 0 && q2.length != 0)
    {
        // Get front nodes and compare them
        var n1 = q1[0];
        var n2 = q2[0];

        if (n1.data != n2.data)
        return false;

        // Remove front nodes from queues
        q1.shift();
        q2.shift();

        /* push left children of both nodes */
        if (n1.left != null && n2.left != null)
        {
            q1.push(n1.left);
            q2.push(n2.left);
        }

        // If one left child is empty and other is not
        else if (n1.left != null || n2.left != null)
            return false;

        // Right child code (Similar to left child code)
        if (n1.right != null && n2.right != null)
        {
            q1.push(n1.right);
            q2.push(n2.right);
        }
        else if (n1.right != null || n2.right != null)
            return false;
    }

    return true;
}

// Utility function to create a new tree node
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Driver code
var root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
var root2 = newNode(1);
root2.left = newNode(2);
root2.right = newNode(3);
root2.left.left = newNode(4);
root2.left.right = newNode(5);
if(areIdentical(root1, root2) == true)
  document.write("Yes");
else
  document.write("No");

</script>

Output:

Yes

Time complexity of above solution is O(n + m) where m and n are number of nodes in two trees.

Space complexity: O(n) space for queue

This article is contributed by Ankur Lathiya .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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