# Check if all levels of two trees are anagrams or not

• Difficulty Level : Medium
• Last Updated : 24 Jun, 2022

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:

1. Write a recursive program for level order traversal of a tree.
2. Traverse each level of both the trees one by one and store the result of traversals in 2 different vectors, one for each tree.
3. 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(nlogn), where n is the number of nodes.

## C++

 `/* Iterative program to check if two trees are level``   ``by level anagram. */``#include ``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 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 q1 = ``new` `LinkedList();``        ``Queue q2 = ``new` `LinkedList();``        ``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 curr_level1 = ``new``                                          ``ArrayList<>();``            ``ArrayList 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`

## C#

 `/* Iterative program to check if two trees``are level by level anagram. */``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{                            ``    ``// A Binary Tree Node``    ``public` `class` `Node``    ``{``        ``public` `Node left, right;``        ``public` `int` `data;``        ``public` `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 q1 = ``new` `Queue();``        ``Queue q2 = ``new` `Queue();``        ``q1.Enqueue(root1);``        ``q2.Enqueue(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.Count, n2 = q2.Count;``    ` `            ``// 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``            ``List<``int``> curr_level1 = ``new` `List<``int``>();``            ``List<``int``> curr_level2 = ``new` `List<``int``>();``            ``while` `(n1 > 0)``            ``{``                ``Node node1 = q1.Peek();``                ``q1.Dequeue();``                ``if` `(node1.left != ``null``)``                    ``q1.Enqueue(node1.left);``                ``if` `(node1.right != ``null``)``                    ``q1.Enqueue(node1.right);``                ``n1--;``    ` `                ``Node node2 = q2.Peek();``                ``q2.Dequeue();``                ``if` `(node2.left != ``null``)``                    ``q2.Enqueue(node2.left);``                ``if` `(node2.right != ``null``)``                    ``q2.Enqueue(node2.right);``    ` `                ``curr_level1.Add(node1.data);``                ``curr_level2.Add(node2.data);``            ``}``    ` `            ``// Check if nodes of current levels are``            ``// anagrams or not.``            ``curr_level1.Sort();``            ``curr_level2.Sort();``            ` `            ``for``(``int` `i = 0;``                    ``i < curr_level1.Count; i++)``            ``if``(curr_level1[i] != curr_level2[i])``                ``return` `false``;``        ``}``        ``return` `true``;``    ``}``    ` `    ``// Driver Code``    ``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);``    ` `        ` `        ``Console.WriteLine(areAnagrams(root1,``                                      ``root2) ?``                                       ``"Yes"` `: ``"No"``);``    ``}``}` `// This code is contributed by Arnab Kundu`

## Javascript

 ``

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.
Efficient Approach:

We can solve the problem in O(n) time complexity by using Hash tables during level order traversal. The idea is to do a level order traversal and in each level check whether the level is an anagram with help of hash tables.

## C++

 `/* Iterative program to check if two trees are level``  ``by level anagram. */``#include ``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 q1, q2;``    ``q1.push(root1);``    ``q2.push(root2);` `    ``// Hashmap to store the elements that occur in each``    ``// level.``    ``unordered_map<``int``, ``int``> m;` `    ``while` `(!q1.empty() && !q2.empty()) {``        ``// 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``        ``while` `(n1--) {``            ``Node* node1 = q1.front();``            ``q1.pop();` `            ``// Insert element into hashmap``            ``m[node1->data]++;` `            ``// Insert left and right nodes into queue if``            ``// exists.``            ``if` `(node1->left != NULL)``                ``q1.push(node1->left);``            ``if` `(node1->right != NULL)``                ``q1.push(node1->right);``        ``}` `        ``while` `(n2--) {``            ``Node* node2 = q2.front();``            ``q2.pop();` `            ``// if element from second tree isn't present in``            ``// the first tree of same level then it can't be``            ``// an anagram.``            ``if` `(m.find(node2->data) == m.end())``                ``return` `false``;` `            ``// Reduce frequency of element if present else``            ``// adds it element to hash map with negative``            ``// frequency.``            ``m[node2->data]--;` `            ``// If frequency of the element becomes zero then``            ``// remove the element from hashmap.``            ``if` `(m[node2->data] == 0)``                ``m.erase(node2->data);` `            ``// Insert left and right nodes into queue if``            ``// exists.``            ``if` `(node2->left != NULL)``                ``q2.push(node2->left);``            ``if` `(node2->right != NULL)``                ``q2.push(node2->right);``        ``}` `        ``// If nodes of current levels are anagrams the``        ``// hashmap wouldn't contain any elements.``        ``if` `(m.size() > 0)``            ``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;``}` `// This code is contributed by Kasina Dheeraj.`

Output

`Yes`

Time complexity: O(N), where N is maximum number of nodes in either of the trees.