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

• Difficulty Level : Easy
• Last Updated : 13 Aug, 2021

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``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 lengths 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;``}`

## Java

 `// Java program to count root to leaf``// paths of different lengths.``import` `java.util.HashMap;``import` `java.util.Map;` `class` `GFG{` `// A binary tree node``static` `class` `Node``{``    ``int` `data;``    ``Node left, right;``};` `// Utility that allocates a new node``// with the given data and null left``// and right pointers.``static` `Node newnode(``int` `data)``{``    ``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.``static` `void` `pathCountUtil(Node node,``        ``HashMap 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``)``    ``{``        ``if` `(!m.containsKey(path_len))``            ``m.put(path_len, ``0``);``            ` `        ``m.put(path_len, m.get(path_len) + ``1``);``        ` `        ``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()``static` `void` `pathCounts(Node root)``{``    ` `    ``// Create an empty hash table``    ``HashMap m = ``new` `HashMap<>();` `    ``// Recursively check in left and right subtrees.``    ``pathCountUtil(root, m, ``1``);``    ` `    ``// Print all path lengths and their counts.``    ``for``(Map.Entry entry : m.entrySet())``    ``{``        ``System.out.printf(``"%d paths have length %d\n"``,``                          ``entry.getValue(),``                          ``entry.getKey());``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``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);``}``}` `// This code is contributed by sanjeev2552`

## 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 lengths 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)`

## C#

 `// C# program to count root to leaf``// paths of different lengths.``using` `System;``using` `System.Collections.Generic;``class` `GFG``{` `  ``// A binary tree node``  ``public``    ``class` `Node``    ``{``      ``public` `        ``int` `data;``      ``public` `        ``Node left,``      ``right;``    ``};` `  ``// Utility that allocates a new node``  ``// with the given data and null left``  ``// and right pointers.``  ``static` `Node newnode(``int` `data)``  ``{``    ``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.``  ``static` `void` `pathCountUtil(Node node,``                            ``Dictionary<``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``)``    ``{``      ``if` `(!m.ContainsKey(path_len))``        ``m.Add(path_len, 1);``      ``else``        ``m[path_len] = m[path_len] + 1;` `      ``return``;``    ``}` `    ``// Recursively call for left and right``    ``// subtrees with path lengths more than 1.``    ``pathCountUtil(node.right, m, path_len + 1);``    ``pathCountUtil(node.left, m, path_len + 1);``  ``}` `  ``// A wrapper over pathCountUtil()``  ``static` `void` `pathCounts(Node root)``  ``{` `    ``// Create an empty hash table``    ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();` `    ``// Recursively check in left and right subtrees.``    ``pathCountUtil(root, m, 1);` `    ``// Print all path lengths and their counts.``    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `m)``    ``{``      ``Console.WriteLine(entry.Value``                        ``+ ``" paths have length "``                        ``+ entry.Key);``    ``}``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``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);``  ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

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

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