# 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 &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 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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