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

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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(n), 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 `

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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