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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<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 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<Integer, Integer> 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<Integer, Integer> m = new HashMap<>();
 
    // Recursively check in left and right subtrees.
    pathCountUtil(root, m, 1);
     
    // Print all path lengths and their counts.
    for(Map.Entry<Integer,
                  Integer> 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




<script>
 
// Javascript program to count root to leaf
// paths of different lengths.
 
// A binary tree node
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};
 
// Utility that allocates a new node
// with the given data and null left
// and right pointers.
function newnode(data)
{
    var 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.
function pathCountUtil(node, m, 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.has(path_len))
            m.set(path_len, 1);
        else
            m.set(path_len, m.get(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()
function pathCounts(root)
{
     
    // Create an empty hash table
    var m = new Map();
     
    // Recursively check in left and right subtrees.
    pathCountUtil(root, m, 1);
     
    // Print all path lengths and their counts.
    m.forEach((value, key) => {
    document.write(value + " paths have length " +
                   key + "<br>"); 
    });
}
 
// Driver code
var 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 itsok
 
</script>

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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