Given two binary trees, we have to check if each of their levels are anagrams of each other or not.
Example:
Tree 1: Level 0 : 1 Level 1 : 3, 2 Level 2 : 5, 4 Tree 2: Level 0 : 1 Level 1 : 2, 3 Level 2 : 4, 5
As we can clearly see all the levels of above two binary trees are anagrams of each other, hence return true.
Naive Approach: Below is the step by step explanation of the naive approach to do this:
- Write a recursive program for level order traversal of a tree.
- Traverse each level of both the trees one by one and store the result of traversals in 2 different vectors, one for each tree.
- Sort both the vectors and compare them iteratively for each level, if they are same for each level then return true else return false.
Time Complexity: O(n^2), where n is the number of nodes.
Efficient Approach:
The idea is based on below article.
Print level order traversal line by line | Set 1
We traverse both trees simultaneously level by level. We store each level both trees in vectors (or array). To check if two vectors are anagram or not, we sort both and then compare.
Time Complexity: O(n), where n is the number of nodes.
C++
/* Iterative program to check if two trees are level by level anagram. */#include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { struct Node *left, *right; int data; }; // Returns true if trees with root1 and root2 // are level by level anagram, else returns false. bool areAnagrams(Node *root1, Node *root2) { // Base Cases if (root1 == NULL && root2 == NULL) return true; if (root1 == NULL || root2 == NULL) return false; // start level order traversal of two trees // using two queues. queue<Node *> q1, q2; q1.push(root1); q2.push(root2); while (1) { // n1 (queue size) indicates number of Nodes // at current level in first tree and n2 indicates // number of nodes in current level of second tree. int n1 = q1.size(), n2 = q2.size(); // If n1 and n2 are different if (n1 != n2) return false; // If level order traversal is over if (n1 == 0) break; // Dequeue all Nodes of current level and // Enqueue all Nodes of next level vector<int> curr_level1, curr_level2; while (n1 > 0) { Node *node1 = q1.front(); q1.pop(); if (node1->left != NULL) q1.push(node1->left); if (node1->right != NULL) q1.push(node1->right); n1--; Node *node2 = q2.front(); q2.pop(); if (node2->left != NULL) q2.push(node2->left); if (node2->right != NULL) q2.push(node2->right); curr_level1.push_back(node1->data); curr_level2.push_back(node2->data); } // Check if nodes of current levels are // anagrams or not. sort(curr_level1.begin(), curr_level1.end()); sort(curr_level2.begin(), curr_level2.end()); if (curr_level1 != curr_level2) 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() { // Constructing both the trees. struct Node* root1 = newNode(1); root1->left = newNode(3); root1->right = newNode(2); root1->right->left = newNode(5); root1->right->right = newNode(4); struct Node* root2 = newNode(1); root2->left = newNode(2); root2->right = newNode(3); root2->left->left = newNode(4); root2->left->right = newNode(5); areAnagrams(root1, root2)? cout << "Yes" : cout << "No"; return 0; } |
Java
/* Iterative program to check if two trees are level by level anagram. */import java.util.ArrayList; import java.util.Collections; import java.util.LinkedList; import java.util.Queue; public class GFG { // A Binary Tree Node static class Node { Node left, right; int data; Node(int data){ this.data = data; left = null; right = null; } } // Returns true if trees with root1 and root2 // are level by level anagram, else returns false. static boolean areAnagrams(Node root1, Node root2) { // Base Cases if (root1 == null && root2 == null) return true; if (root1 == null || root2 == null) return false; // start level order traversal of two trees // using two queues. Queue<Node> q1 = new LinkedList<Node>(); Queue<Node> q2 = new LinkedList<Node>(); q1.add(root1); q2.add(root2); while (true) { // n1 (queue size) indicates number of // Nodes at current level in first tree // and n2 indicates number of nodes in // current level of second tree. int n1 = q1.size(), n2 = q2.size(); // If n1 and n2 are different if (n1 != n2) return false; // If level order traversal is over if (n1 == 0) break; // Dequeue all Nodes of current level and // Enqueue all Nodes of next level ArrayList<Integer> curr_level1 = new ArrayList<>(); ArrayList<Integer> curr_level2 = new ArrayList<>(); while (n1 > 0) { Node node1 = q1.peek(); q1.remove(); if (node1.left != null) q1.add(node1.left); if (node1.right != null) q1.add(node1.right); n1--; Node node2 = q2.peek(); q2.remove(); if (node2.left != null) q2.add(node2.left); if (node2.right != null) q2.add(node2.right); curr_level1.add(node1.data); curr_level2.add(node2.data); } // Check if nodes of current levels are // anagrams or not. Collections.sort(curr_level1); Collections.sort(curr_level2); if (!curr_level1.equals(curr_level2)) return false; } return true; } // Driver program to test above functions public static void main(String args[]) { // Constructing both the trees. Node root1 = new Node(1); root1.left = new Node(3); root1.right = new Node(2); root1.right.left = new Node(5); root1.right.right = new Node(4); 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); System.out.println(areAnagrams(root1, root2)? "Yes" : "No"); } } // This code is contributed by Sumit Ghosh |
Python3
# Iterative program to check if two # trees are level by level anagram # A Binary Tree Node # Utility function to create a # new tree Node class newNode: def __init__(self, data): self.data = data self.left = self.right = None # Returns true if trees with root1 # and root2 are level by level # anagram, else returns false. def areAnagrams(root1, root2) : # Base Cases if (root1 == None and root2 == None) : return True if (root1 == None or root2 == None) : return False # start level order traversal of # two trees using two queues. q1 = [] q2 = [] q1.append(root1) q2.append(root2) while (1) : # n1 (queue size) indicates number # of Nodes at current level in first # tree and n2 indicates number of nodes # in current level of second tree. n1 = len(q1) n2 = len(q2) # If n1 and n2 are different if (n1 != n2): return False # If level order traversal is over if (n1 == 0): break # Dequeue all Nodes of current level # and Enqueue all Nodes of next level curr_level1 = [] curr_level2 = [] while (n1 > 0): node1 = q1[0] q1.pop(0) if (node1.left != None) : q1.append(node1.left) if (node1.right != None) : q1.append(node1.right) n1 -= 1 node2 = q2[0] q2.pop(0) if (node2.left != None) : q2.append(node2.left) if (node2.right != None) : q2.append(node2.right) curr_level1.append(node1.data) curr_level2.append(node2.data) # Check if nodes of current levels # are anagrams or not. curr_level1.sort() curr_level2.sort() if (curr_level1 != curr_level2) : return False return True # Driver Code if __name__ == '__main__': # Constructing both the trees. root1 = newNode(1) root1.left = newNode(3) root1.right = newNode(2) root1.right.left = newNode(5) root1.right.right = newNode(4) root2 = newNode(1) root2.left = newNode(2) root2.right = newNode(3) root2.left.left = newNode(4) root2.left.right = newNode(5) if areAnagrams(root1, root2): print("Yes") else: print("No") # This code is contributed # by SHUBHAMSINGH10 |
Output:
Yes
Note: In the above program we are comparing the vectors storing each level of a tree directly using not equal to function ‘ != ‘ which compares the vectors first on the basis of their size and then on the basis of their content, hence saving our work of iteratively comparing the vectors.
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