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; } |
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 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 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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