Given two Binary Trees we have to detect if the two trees are Isomorphic. Two trees are called isomorphic if one of them can be obtained from another by a series of flips, i.e. by swapping left and right children of a number of nodes. Any number of nodes at any level can have their children swapped.
Note: Two empty trees are isomorphic.
For example, the following two trees are isomorphic with the following sub-trees flipped: 2 and 3, NULL and 6, 7, and 8.
Approach:
To solve the question mentioned above we traverse both the trees iteratively using level order traversal and store the levels in a queue data structure. There are following two conditions at each level:
- The value of nodes has to be the same.
- The number of nodes at each level should be the same.
Check the size of the queue to match the second condition mentioned above. Store the nodes of each level of the first tree as key with a value and for the second tree, we will store all the nodes of a level in a vector. If the key is found we will decrease the value to keep track of how many nodes with the same value exists at a level. If the value becomes zero that means the first tree has only this much number of nodes, and we will remove it as a key. At the end of each level, we will iterate through the array and check whether each value exists on the map or not. There will be three conditions:
- If the key is not found then the first tree doesn’t contain a node found in the second tree at the same level.
- If the key is found but the value becomes negative then the second tree has more nodes with the same value as the first tree.
- If map size is not zero, this means there are some keys still left which means the first tree has a node that doesn’t match with any node in the second tree.
Below is the implementation of the above approach:
C++
// C++ program to find two// Binary Tree are Isomorphic or not#include <bits/stdc++.h>using namespace std;/* A binary tree node has data,pointer to left and right children */struct node { int data; struct node* left; struct node* right;};/* function to return if tree are isomorphic or not*/bool isIsomorphic(node* root1, node* root2){ // if Both roots are null // then tree is isomorphic if (root1 == NULL and root2 == NULL) return true; // check if one node is false else if (root1 == NULL or root2 == NULL) return false; queue<node *> q1, q2; // enqueue roots q1.push(root1); q2.push(root2); int level = 0; int size; vector<int> v2; unordered_map<int, int> mp; while (!q1.empty() and !q2.empty()) { // check if no. of nodes are // not same at a given level if (q1.size() != q2.size()) return false; size = q1.size(); level++; v2.clear(); mp.clear(); while (size--) { node* temp1 = q1.front(); node* temp2 = q2.front(); // dequeue the nodes q1.pop(); q2.pop(); // check if value // exists in the map if (mp.find(temp1->data) == mp.end()) mp[temp1->data] = 1; else mp[temp1->data]++; v2.push_back(temp2->data); // enqueue the child nodes if (temp1->left) q1.push(temp1->left); if (temp1->right) q1.push(temp1->right); if (temp2->left) q2.push(temp2->left); if (temp2->right) q2.push(temp2->right); } // Iterate through each node at a level // to check whether it exists or not. for (auto i : v2) { if (mp.find(i) == mp.end()) return false; else { mp[i]--; if (mp[i] < 0) return false; else if (mp[i] == 0) mp.erase(i); } } // check if the key remain if (mp.size() != 0) return false; } return true;}/* function that allocates a new node with the given data and NULL left and right pointers. */node* newnode(int data){ node* temp = new node; temp->data = data; temp->left = NULL; temp->right = NULL; return (temp);}/* Driver program*/int main(){ // create tree struct node* n1 = newnode(1); n1->left = newnode(2); n1->right = newnode(3); n1->left->left = newnode(4); n1->left->right = newnode(5); n1->right->left = newnode(6); n1->left->right->left = newnode(7); n1->left->right->right = newnode(8); struct node* n2 = newnode(1); n2->left = newnode(3); n2->right = newnode(2); n2->right->left = newnode(4); n2->right->right = newnode(5); n2->left->right = newnode(6); n2->right->right->left = newnode(8); n2->right->right->right = newnode(7); if (isIsomorphic(n1, n2) == true) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to find two// Binary Tree are Isomorphic or notimport java.util.ArrayList;import java.util.HashMap;import java.util.LinkedList;import java.util.Queue;class GFG{// A binary tree node has data, // pointer to left and right childrenstatic class Node { int data; Node left, right; public Node(int data) { this.data = data; this.left = this.right = null; }};// Function to return if tree // are isomorphic or notstatic boolean isIsomorphic(Node root1, Node root2) { // If Both roots are null // then tree is isomorphic if (root1 == null && root2 == null) return true; // Check if one node // is false else if (root1 == null || root2 == null) return false; Queue<Node> q1 = new LinkedList<>(), q2 = new LinkedList<>(); // Enqueue roots q1.add(root1); q2.add(root2); int level = 0; int size; ArrayList<Integer> v2 = new ArrayList<>(); HashMap<Integer, Integer> mp = new HashMap<>(); while (!q1.isEmpty() && !q2.isEmpty()) { // check if no. of nodes are // not same at a given level if (q1.size() != q2.size()) return false; size = q1.size(); level++; v2.clear(); mp.clear(); while (size-- > 0) { // Dequeue the nodes Node temp1 = q1.poll(); Node temp2 = q2.poll(); // Check if value // exists in the map if (!mp.containsKey(temp1.data)) mp.put(temp1.data, 1); else mp.put(temp1.data, mp.get(temp1.data) + 1); v2.add(temp2.data); // Enqueue the child nodes if (temp1.left != null) q1.add(temp1.left); if (temp1.right != null) q1.add(temp1.right); if (temp2.left != null) q2.add(temp2.left); if (temp2.right != null) q2.add(temp2.right); } // Iterate through each node // at a level to check whether // it exists or not. for (Integer i : v2) { if (!mp.containsKey(i)) return false; else { mp.put(i, mp.get(i) - 1); if (mp.get(i) < 0) return false; else if (mp.get(i) == 0) mp.remove(i); } } // Check if the key remain if (mp.size() != 0) return false; } return true;}// Driver programpublic static void main(String[] args) { // Create tree Node n1 = new Node(1); n1.left = new Node(2); n1.right = new Node(3); n1.left.left = new Node(4); n1.left.right = new Node(5); n1.right.left = new Node(6); n1.left.right.left = new Node(7); n1.left.right.right = new Node(8); Node n2 = new Node(1); n2.left = new Node(3); n2.right = new Node(2); n2.right.left = new Node(4); n2.right.right = new Node(5); n2.left.right = new Node(6); n2.right.right.left = new Node(8); n2.right.right.right = new Node(7); if (isIsomorphic(n1, n2)) System.out.println("Yes"); else System.out.println("No");} }// This code is contributed by sanjeev2552 |
Python3
# Python3 program to find two# Binary Tree are Isomorphic or notfrom collections import dequeclass node: def __init__(self, x): self.data = x self.left = None self.right = None# Function to return if tree are# isomorphic or notdef isIsomorphic(root1, root2): # If Both roots are null # then tree is isomorphic if (root1 == None and root2 == None): return True # Check if one node is false elif (root1 == None or root2 == None): return False q1 = deque() q2 = deque() # enqueue roots q1.append(root1) q2.append(root2) level = 0 size = 0 v1 = [] m2 = {} while (len(q1) > 0 and len(q2) > 0): # Check if no. of nodes are # not same at a given level if (len(q1) != len(q2)): return False size = len(q1) level += 1 v2 = [] mp = {} while (size): temp1 = q1.popleft() temp2 = q2.popleft() # Check if value # exists in the map mp[temp1.data] = mp.get(temp1.data, 0) + 1 v2.append(temp2.data) # enqueue the child nodes if (temp1.left): q1.append(temp1.left) if (temp1.right): q1.append(temp1.right) if (temp2.left): q2.append(temp2.left) if (temp2.right): q2.append(temp2.right) v1 = v2 m2 = mp size -= 1 # Iterate through each node at a level # to check whether it exists or not. for i in v1: if i in m2: return True else: m2[i] -= 1 if (m2[i] < 0): return False elif (m2[i] == 0): del m2[i] # Check if the key remain if (len(m2) == 0): return False return True# Driver codeif __name__ == '__main__': # Create tree n1 = node(1) n1.left = node(2) n1.right = node(3) n1.left.left = node(4) n1.left.right = node(5) n1.right.left = node(6) n1.left.right.left = node(7) n1.left.right.right = node(8) n2 = node(1) n2.left = node(3) n2.right = node(2) n2.right.left = node(4) n2.right.right = node(5) n2.left.right = node(6) n2.right.right.left = node(8) n2.right.right.right = node(7) if (isIsomorphic(n1, n2) == True): print("Yes") else: print("No")# This code is contributed by mohit kumar 29 |
C#
// C# program to find two// Binary Tree are Isomorphic or notusing System;using System.Collections;using System.Collections.Generic; class GFG{ // A binary tree node has data, // pointer to left and right childrenclass Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = this.right = null; }}; // Function to return if tree // are isomorphic or notstatic bool isIsomorphic(Node root1, Node root2) { // If Both roots are null // then tree is isomorphic if (root1 == null && root2 == null) return true; // Check if one node // is false else if (root1 == null || root2 == null) return false; Queue q1 = new Queue(); Queue q2 = new Queue(); // roots q1.Enqueue(root1); q2.Enqueue(root2); int level = 0; int size; ArrayList v2 = new ArrayList(); Dictionary<int,int> mp = new Dictionary<int,int>(); while (q1.Count != 0 && q2.Count != 0) { // check if no. of nodes are // not same at a given level if (q1.Count != q2.Count) return false; size = q1.Count; level++; v2.Clear(); mp.Clear(); while (size-- > 0) { // Dequeue the nodes Node temp1 = (Node)q1.Dequeue(); Node temp2 = (Node)q2.Dequeue(); // Check if value // exists in the map if (!mp.ContainsKey(temp1.data)) mp[temp1.data] = 1; else mp[temp1.data]++; v2.Add(temp2.data); // the child nodes if (temp1.left != null) q1.Enqueue(temp1.left); if (temp1.right != null) q1.Enqueue(temp1.right); if (temp2.left != null) q2.Enqueue(temp2.left); if (temp2.right != null) q2.Enqueue(temp2.right); } // Iterate through each node // at a level to check whether // it exists or not. foreach (int i in v2) { if (!mp.ContainsKey(i)) return false; else { mp[i]--; if (mp[i] < 0) return false; else if (mp[i] == 0) mp.Remove(i); } } // Check if the key remain if (mp.Count != 0) return false; } return true;} // Driver programpublic static void Main(string[] args) { // Create tree Node n1 = new Node(1); n1.left = new Node(2); n1.right = new Node(3); n1.left.left = new Node(4); n1.left.right = new Node(5); n1.right.left = new Node(6); n1.left.right.left = new Node(7); n1.left.right.right = new Node(8); Node n2 = new Node(1); n2.left = new Node(3); n2.right = new Node(2); n2.right.left = new Node(4); n2.right.right = new Node(5); n2.left.right = new Node(6); n2.right.right.left = new Node(8); n2.right.right.right = new Node(7); if (isIsomorphic(n1, n2)) Console.WriteLine("Yes"); else Console.WriteLine("No");} }// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript program to find two// Binary Tree are Isomorphic or not// A binary tree node has data,// pointer to left and right childrenclass Node{ // Utility function to // create a new node constructor(key) { this.data = key; this.left = this.right = null; }}// Function to return if tree// are isomorphic or notfunction isIsomorphic(root1, root2){ // If Both roots are null // then tree is isomorphic if (root1 == null && root2 == null) return true; // Check if one node // is false else if (root1 == null || root2 == null) return false; let q1 =[], q2 = []; // Enqueue roots q1.push(root1); q2.push(root2); let level = 0; let size; let v2 = []; let mp = new Map(); while (q1.length != 0 && q2.length != 0) { // Check if no. of nodes are // not same at a given level if (q1.length != q2.length) return false; size = q1.length; level++; v2 = []; while (size-- > 0) { // Dequeue the nodes let temp1 = q1.shift(); let temp2 = q2.shift(); // Check if value // exists in the map if (!mp.has(temp1.data)) mp.set(temp1.data, 1); else mp.set(temp1.data, mp.get(temp1.data) + 1); v2.push(temp2.data); // Enqueue the child nodes if (temp1.left != null) q1.push(temp1.left); if (temp1.right != null) q1.push(temp1.right); if (temp2.left != null) q2.push(temp2.left); if (temp2.right != null) q2.push(temp2.right); } // Iterate through each node // at a level to check whether // it exists or not. for(let i = 0; i < v2.length; i++) { if (!mp.has(v2[i])) return false; else { mp.set(v2[i], mp.get(v2[i]) - 1); if (mp.get(v2[i]) < 0) return false; else if (mp.get(v2[i]) == 0) mp.delete(v2[i]); } } // Check if the key remain if (mp.size != 0) return false; } return true;}// Driver code// Create treelet n1 = new Node(1);n1.left = new Node(2);n1.right = new Node(3);n1.left.left = new Node(4);n1.left.right = new Node(5);n1.right.left = new Node(6);n1.left.right.left = new Node(7);n1.left.right.right = new Node(8);let n2 = new Node(1);n2.left = new Node(3);n2.right = new Node(2);n2.right.left = new Node(4);n2.right.right = new Node(5);n2.left.right = new Node(6);n2.right.right.left = new Node(8);n2.right.right.right = new Node(7);if (isIsomorphic(n1, n2)) document.write("Yes");else document.write("No");// This code is contributed by patel2127</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
