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
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 caseint 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 lenghts 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 codepublic 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 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) |
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 lenghts 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 |
Output:
1 paths have length 4 2 paths have length 3
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