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Iterative function to check if two trees are identical

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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 == nullreturn 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
 Using Iterative Post-Order Traversal and Two Stacks:

The basic idea behind this approach is to traverse both trees in a postorder fashion iteratively, and compare their nodes one by one. We can use two stacks to do this. We start with pushing the root nodes of both trees onto their respective stacks. 

Follow the steps to implement the idea:

  1. we repeat the following steps until both stacks are empty:
  2. Pop the top node from each stack.
  3. Compare the popped nodes. If they are not identical, return false.
  4. Push the right subtree of both nodes (if they exist) onto their respective stacks.
  5. Push the left subtree of both nodes (if they exist) onto their respective stacks.

Below is the implementation of the above approach:

C++

// C++ code to implement Iterative Postorder Traversal using
// two stacks
#include <iostream>
#include <stack>
using namespace std;
 
/* A binary tree node */
struct Node {
    int data;
    Node *left, *right;
    Node(int x)
    {
        data = x;
        left = right = NULL;
    }
};
 
/* Iterative Postorder Traversal to check if two binary
 * trees are identical */
bool isIdentical(Node* r1, Node* r2)
{
    stack<Node*> stack1, stack2;
 
    // loop until both trees are completely traversed
    while (r1 != NULL || !stack1.empty() || r2 != NULL
           || !stack2.empty()) {
        // push all left nodes of first tree in stack1
        while (r1 != NULL) {
            stack1.push(r1);
            r1 = r1->left;
        }
        // push all left nodes of second tree in stack2
        while (r2 != NULL) {
            stack2.push(r2);
            r2 = r2->left;
        }
        // if size of both stacks is different, trees are
        // not identical
        if (stack1.size() != stack2.size())
            return false;
        // pop one node from each stack and compare their
        // data
        r1 = stack1.top();
        stack1.pop();
        r2 = stack2.top();
        stack2.pop();
        if (r1->data != r2->data)
            return false;
        // move to the right of the popped nodes
        r1 = r1->right;
        r2 = r2->right;
    }
    // both trees are identical
    return true;
}
 
/* Driver code */
int main()
{
    // Construct the first tree
    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);
 
    // Construct the second tree
    Node* root2 = new Node(1);
    root2->left = new Node(2);
    root2->right = new Node(3);
    root2->left->left = new Node(4);
    root2->left->right = new Node(5);
 
    // Check if the trees are identical
    if (isIdentical(root1, root2))
        cout << "Both trees are identical";
    else
        cout << "Both trees are not identical";
 
    return 0;
}
// This code is contributed by Veerendra_Singh_Rajpoot

                    

Java

import java.util.Stack;
 
// A Java program to implement Iterative Postorder Traversal using two stacks
 
// A binary tree node
class Node {
    int data;
    Node left, right;
     
    // Constructor
    Node(int x) {
        data = x;
        left = right = null;
    }
}
 
public class IdenticalBinaryTrees {
    // Iterative Postorder Traversal to check if two binary trees are identical
    static boolean isIdentical(Node r1, Node r2) {
        Stack<Node> stack1 = new Stack<>();
        Stack<Node> stack2 = new Stack<>();
         
        // loop until both trees are completely traversed
        while (r1 != null || !stack1.empty() || r2 != null || !stack2.empty()) {
            // push all left nodes of the first tree in stack1
            while (r1 != null) {
                stack1.push(r1);
                r1 = r1.left;
            }
            // push all left nodes of the second tree in stack2
            while (r2 != null) {
                stack2.push(r2);
                r2 = r2.left;
            }
            // if the size of both stacks is different, trees are not identical
            if (stack1.size() != stack2.size())
                return false;
            // pop one node from each stack and compare their data
            r1 = stack1.pop();
            r2 = stack2.pop();
            if (r1.data != r2.data)
                return false;
            // move to the right of the popped nodes
            r1 = r1.right;
            r2 = r2.right;
        }
        // both trees are identical
        return true;
    }
//Driver code
    public static void main(String[] args) {
        // Construct the first tree
        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);
 
        // Construct the second tree
        Node root2 = new Node(1);
        root2.left = new Node(2);
        root2.right = new Node(3);
        root2.left.left = new Node(4);
        root2.left.right = new Node(5);
 
        // Check if the trees are identical
        if (isIdentical(root1, root2))
            System.out.println("Both trees are identical");
        else
            System.out.println("Both trees are not identical");
    }
}

                    

Python3

# Python code to implement Iterative Postorder Traversal using
# two stacks
 
# A class representing a node in the binary tree
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Iterative Postorder Traversal to check if two binary
# trees are identical
def isIdentical(r1, r2):
    stack1 = []
    stack2 = []
 
    # Loop until both trees are completely traversed
    while r1 or stack1 or r2 or stack2:
        # Push all left nodes of the first tree in stack1
        while r1:
            stack1.append(r1)
            r1 = r1.left
 
        # Push all left nodes of the second tree in stack2
        while r2:
            stack2.append(r2)
            r2 = r2.left
 
        # If size of both stacks is different, trees are not identical
        if len(stack1) != len(stack2):
            return False
 
        # Pop one node from each stack and compare their data
        r1 = stack1.pop()
        r2 = stack2.pop()
 
        if r1.data != r2.data:
            return False
 
        # Move to the right of the popped nodes
        r1 = r1.right
        r2 = r2.right
 
    # Both trees are identical
    return True
 
# Driver code
if __name__ == '__main__':
    # Construct the first tree
    root1 = Node(1)
    root1.left = Node(2)
    root1.right = Node(3)
    root1.left.left = Node(4)
    root1.left.right = Node(5)
 
    # Construct the second tree
    root2 = Node(1)
    root2.left = Node(2)
    root2.right = Node(3)
    root2.left.left = Node(4)
    root2.left.right = Node(5)
 
    # Check if the trees are identical
    if isIdentical(root1, root2):
        print("Both trees are identical")
    else:
        print("Both trees are not identical")

                    

C#

// C# code to implement Iterative Postorder Traversal using
// two stacks
using System;
using System.Collections.Generic;
 
// A binary tree node
class Node
{
    public int data;
    public Node left, right;
    public Node(int x)
    {
        data = x;
        left = right = null;
    }
}
 
class IterativePostorderTraversal
{
    static bool IsIdentical(Node r1, Node r2)
    {
        Stack<Node> stack1 = new Stack<Node>();
        Stack<Node> stack2 = new Stack<Node>();
 
        // Loop until both trees are completely traversed
        while (r1 != null || stack1.Count > 0 || r2 != null || stack2.Count > 0)
        {
            // Push all left nodes of the first tree in stack1
            while (r1 != null)
            {
                stack1.Push(r1);
                r1 = r1.left;
            }
            // Push all left nodes of the second tree in stack2
            while (r2 != null)
            {
                stack2.Push(r2);
                r2 = r2.left;
            }
            // If the sizes of both stacks are different, trees are not identical
            if (stack1.Count != stack2.Count)
                return false;
            // Pop one node from each stack and compare their data
            r1 = stack1.Pop();
            r2 = stack2.Pop();
            if (r1.data != r2.data)
                return false;
            // Move to the right of the popped nodes
            r1 = r1.right;
            r2 = r2.right;
        }
        // Both trees are identical
        return true;
    }
// Driver code
    static void Main(string[] args)
    {
        // Construct the first tree
        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);
 
        // Construct the second tree
        Node root2 = new Node(1);
        root2.left = new Node(2);
        root2.right = new Node(3);
        root2.left.left = new Node(4);
        root2.left.right = new Node(5);
 
        // Check if the trees are identical
        if (IsIdentical(root1, root2))
            Console.WriteLine("Both trees are identical");
        else
            Console.WriteLine("Both trees are not identical");
    }
}

                    

Javascript

// Define a binary tree node structure
class Node {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// Function to perform iterative postorder traversal and check if two binary trees are identical
function isIdentical(r1, r2) {
    const stack1 = [];
    const stack2 = [];
 
    // Loop until both trees are completely traversed
    while (r1 || stack1.length > 0 || r2 || stack2.length > 0) {
        // Push all left nodes of the first tree into stack1
        while (r1) {
            stack1.push(r1);
            r1 = r1.left;
        }
 
        // Push all left nodes of the second tree into stack2
        while (r2) {
            stack2.push(r2);
            r2 = r2.left;
        }
 
        // If the sizes of both stacks are different, the trees are not identical
        if (stack1.length !== stack2.length)
            return false;
 
        // Pop one node from each stack and compare their data
        r1 = stack1.pop();
        r2 = stack2.pop();
 
        if (r1.data !== r2.data)
            return false;
 
        // Move to the right of the popped nodes
        r1 = r1.right;
        r2 = r2.right;
    }
 
    // Both trees are identical
    return true;
}
 
// Driver code
function main() {
    // Construct the first tree
    const 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);
 
    // Construct the second tree
    const root2 = new Node(1);
    root2.left = new Node(2);
    root2.right = new Node(3);
    root2.left.left = new Node(4);
    root2.left.right = new Node(5);
 
    // Check if the trees are identical
    if (isIdentical(root1, root2))
        console.log("Both trees are identical");
    else
        console.log("Both trees are not identical");
}
 
// Run the main function
main();

                    

Output
Both trees are identical







Time Complexity: O(N) , The time complexity of the isIdentical function is O(n), where n is the total number of nodes in the trees. This is because we visit each node exactly once and perform a constant amount of work at each node.

Auxiliary Space: O(H) , The space complexity of the isIdentical function is O(h), where h is the maximum height of the two trees. This is because we use two stacks to keep track of the nodes in the two trees
This article is contributed by Ankur Lathiya .

 



Last Updated : 11 Oct, 2023
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