# Root to leaf paths having equal lengths in a Binary Tree

Given a binary tree, print the number of root to leaf paths having equal lengths.

Examples:

```Input : Root of below tree
10
/   \
8      2
/  \    /  \
3     5  2    4
Output : 4 paths are of length 3.

Input : Root of below tree
10
/   \
8      2
/  \    /  \
3    5  2    4
/               \
9                 1
Output : 2 paths are of length 3
2 paths are of length 4
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to traverse the tree and keep track of path length. Whenever we reach a leaf node, we increment path length count in a hash map.
Once we have traverse the tree, hash map has counts of distinct path lengths. Finally we print contents of hash map.

## C++

 `// C++ program to count root to leaf paths of different ` `// lengths. ` `#include ` `using` `namespace` `std; ` ` `  `/* A binary tree node */` `struct` `Node ` `{ ` `    ``int` `data; ` `    ``struct` `Node* left, *right; ` `}; ` ` `  `/* utility that allocates a new node with the ` `   ``given data and NULL left and right pointers. */` `struct` `Node* newnode(``int` `data) ` `{ ` `    ``struct` `Node* node = ``new` `Node; ` `    ``node->data = data; ` `    ``node->left = node->right  = NULL; ` `    ``return` `(node); ` `} ` ` `  `// Function to store counts of different root to leaf ` `// path lengths in hash map m. ` `void` `pathCountUtil(Node *node, unordered_map<``int``, ``int``> &m, ` `                                             ``int` `path_len) ` `{ ` `    ``// Base condition ` `    ``if` `(node == NULL) ` `        ``return``; ` ` `  `    ``// If leaf node reached, increment count of path ` `    ``// length of this root to leaf path. ` `    ``if` `(node->left == NULL && node->right == NULL) ` `    ``{ ` `         ``m[path_len]++; ` `         ``return``; ` `    ``} ` ` `  `    ``// Recursively call for left and right subtrees with ` `    ``// path lengths more than 1. ` `    ``pathCountUtil(node->left, m, path_len+1); ` `    ``pathCountUtil(node->right, m, path_len+1); ` `} ` ` `  `// A wrapper over pathCountUtil() ` `void` `pathCounts(Node *root) ` `{ ` `   ``// create an empty hash table ` `   ``unordered_map<``int``, ``int``> m; ` ` `  `   ``// Recursively check in left and right subtrees. ` `   ``pathCountUtil(root, m, 1); ` ` `  `   ``// Print all path lenghts and their counts. ` `   ``for` `(``auto` `itr=m.begin(); itr != m.end(); itr++) ` `      ``cout << itr->second << ``" paths have length "` `           ``<< itr->first << endl; ` `} ` ` `  `// Driver program to run the case ` `int` `main() ` `{ ` `    ``struct` `Node *root = newnode(8); ` `    ``root->left    = newnode(5); ` `    ``root->right   = newnode(4); ` `    ``root->left->left = newnode(9); ` `    ``root->left->right = newnode(7); ` `    ``root->right->right = newnode(11); ` `    ``root->right->right->left = newnode(3); ` `    ``pathCounts(root); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program to count root to leaf  ` `# paths of different lengths. ` `     `  `# Binary Tree Node  ` `""" utility that allocates a newNode  ` `with the given key """` `class` `newnode:  ` ` `  `    ``# Construct to create a newNode  ` `    ``def` `__init__(``self``, key):  ` `        ``self``.key ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `         `  `# Function to store counts of different  ` `# root to leaf path lengths in hash map m.  ` `def` `pathCountUtil(node, m,path_len) : ` ` `  `    ``# Base condition  ` `    ``if` `(node ``=``=` `None``) : ` `        ``return` ` `  `    ``# If leaf node reached, increment count of  ` `    ``# path length of this root to leaf path.  ` `    ``if` `(node.left ``=``=` `None` `and` `node.right ``=``=` `None``):      ` `        ``if` `path_len[``0``] ``not` `in` `m: ` `            ``m[path_len[``0``]] ``=` `0` `        ``m[path_len[``0``]] ``+``=` `1` `        ``return` ` `  `    ``# Recursively call for left and right  ` `    ``# subtrees with path lengths more than 1. ` `    ``pathCountUtil(node.left, m, [path_len[``0``] ``+` `1``]) ` `    ``pathCountUtil(node.right, m, [path_len[``0``] ``+` `1``])  ` ` `  `# A wrapper over pathCountUtil()  ` `def` `pathCounts(root) : ` ` `  `    ``# create an empty hash table  ` `    ``m ``=` `{} ` `    ``path_len ``=` `[``1``] ` `     `  `    ``# Recursively check in left and right subtrees.  ` `    ``pathCountUtil(root, m, path_len)  ` ` `  `    ``# Print all path lenghts and their counts.  ` `    ``for` `itr ``in` `sorted``(m, reverse ``=` `True``): ` `        ``print``(m[itr], ``" paths have length "``, itr)  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``root ``=` `newnode(``8``)  ` `    ``root.left ``=` `newnode(``5``)  ` `    ``root.right ``=` `newnode(``4``)  ` `    ``root.left.left ``=` `newnode(``9``)  ` `    ``root.left.right ``=` `newnode(``7``)  ` `    ``root.right.right ``=` `newnode(``11``)  ` `    ``root.right.right.left ``=` `newnode(``3``)  ` `    ``pathCounts(root) ` ` `  `# This code is contributed by ` `# Shubham Singh(SHUBHAMSINGH10) `

Output:

```1 paths have length 4
2 paths have length 3
```

This article is contributed by Sahil Chhabra (KILLER). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : SHUBHAMSINGH10

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