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

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<bits/stdc++.h> ` `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; ` `} ` |

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## 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) ` |

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**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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