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> |
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:
- we repeat the following steps until both stacks are empty:
- Pop the top node from each stack.
- Compare the popped nodes. If they are not identical, return false.
- Push the right subtree of both nodes (if they exist) onto their respective stacks.
- 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(); |
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 .