Maximum distinct nodes in a Root to leaf path
Given a Binary Tree, find count of distinct nodes in all the root to leaf paths and print the maximum.
Examples:
Input : 1
/ \
2 3
/ \ / \
4 5 6 3
\ \
8 9
Output : 4
The root to leaf path with maximum distinct
nodes is 1-3-6-8.
A simple solution is to explore all root to leaf paths. In every root to leaf path, count distinct nodes and finally return the maximum count.
An efficient solution is to use hashing. We recursively traverse the tree and maintain count of distinct nodes on path from root to current node. We recur for left and right subtrees and finally return maximum of two values.
Below is implementation of above idea
C++
// C++ program to find count of distinct nodes// on a path with maximum distinct nodes.#include <bits/stdc++.h>using namespace std;// A node of binary treestruct Node { int data; struct Node *left, *right;};// A utility function to create a new Binary// Tree nodeNode* newNode(int data){ Node* temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp;}int largestUinquePathUtil(Node* node, unordered_map<int, int> m){ if (!node) return m.size(); // put this node into hash m[node->data]++; int max_path = max(largestUinquePathUtil(node->left, m), largestUinquePathUtil(node->right, m)); // remove current node from path "hash" m[node->data]--; // if we reached a condition where all duplicate value // of current node is deleted if (m[node->data] == 0) m.erase(node->data); return max_path;}// A utility function to find long unique value pathint largestUinquePath(Node* node){ if (!node) return 0; // hash that store all node value unordered_map<int, int> hash; // return max length unique value path return largestUinquePathUtil(node, hash);}// Driver program to test above functionsint main(){ // Create binary tree shown in above figure Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); root->right->left->right = newNode(8); root->right->right->right = newNode(9); cout << largestUinquePath(root) << endl; return 0;} |
Java
// Java program to find count of distinct nodes// on a path with maximum distinct nodes.import java.util.*;class GFG{// A node of binary treestatic class Node{ int data; Node left, right;};// A utility function to create a new Binary// Tree nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}static int largestUinquePathUtil(Node node, HashMap<Integer, Integer> m){ if (node == null) return m.size(); // put this node into hash if(m.containsKey(node.data)) { m.put(node.data, m.get(node.data) + 1); } else { m.put(node.data, 1); } int max_path = Math.max(largestUinquePathUtil(node.left, m), largestUinquePathUtil(node.right, m)); // remove current node from path "hash" if(m.containsKey(node.data)) { m.put(node.data, m.get(node.data) - 1); } // if we reached a condition where all duplicate value // of current node is deleted if (m.get(node.data) == 0) m.remove(node.data); return max_path;}// A utility function to find long unique value pathstatic int largestUinquePath(Node node){ if (node == null) return 0; // hash that store all node value HashMap<Integer, Integer> hash = new HashMap<Integer, Integer>(); // return max length unique value path return largestUinquePathUtil(node, hash);}// Driver Codepublic static void main(String[] args){ // Create binary tree shown in above figure Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.right.left.right = newNode(8); root.right.right.right = newNode(9); System.out.println(largestUinquePath(root)); }}// This code is contributed by Princi Singh |
Python3
# Python3 program to find count of# distinct nodes on a path with# maximum distinct nodes.# A utility class to create a# new Binary Tree nodeclass newNode: def __init__(self, data): self.data = data self.left = self.right = Nonedef largestUinquePathUtil(node, m): if (not node): return len(m) # put this node into hash if node.data in m: m[node.data] += 1 else: m[node.data] = 1 max_path = max(largestUinquePathUtil(node.left, m), largestUinquePathUtil(node.right, m)) # remove current node from path "hash" m[node.data] -= 1 # if we reached a condition # where all duplicate value # of current node is deleted if (m[node.data] == 0): del m[node.data] return max_path# A utility function to find# long unique value pathdef largestUinquePath(node): if (not node): return 0 # hash that store all node value Hash = {} # return max length unique value path return largestUinquePathUtil(node, Hash)# Driver Codeif __name__ == '__main__': # Create binary tree shown # in above figure root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) root.right.left = newNode(6) root.right.right = newNode(7) root.right.left.right = newNode(8) root.right.right.right = newNode(9) print(largestUinquePath(root))# This code is contributed by PranchalK |
C#
// C# program to find count of distinct nodes// on a path with maximum distinct nodes.using System;using System.Collections.Generic;class GFG{// A node of binary treepublic class Node{ public int data; public Node left, right;};// A utility function to create a new Binary// Tree nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.left = temp.right = null; return temp;}static int largestUinquePathUtil(Node node, Dictionary<int, int> m){ if (node == null) return m.Count; // put this node into hash if(m.ContainsKey(node.data)) { m[node.data] = m[node.data] + 1; } else { m.Add(node.data, 1); } int max_path = Math.Max(largestUinquePathUtil(node.left, m), largestUinquePathUtil(node.right, m)); // remove current node from path "hash" if(m.ContainsKey(node.data)) { m[node.data] = m[node.data] - 1; } // if we reached a condition where all // duplicate value of current node is deleted if (m[node.data] == 0) m.Remove(node.data); return max_path;}// A utility function to find long unique value pathstatic int largestUinquePath(Node node){ if (node == null) return 0; // hash that store all node value Dictionary<int, int> hash = new Dictionary<int, int>(); // return max length unique value path return largestUinquePathUtil(node, hash);}// Driver Codepublic static void Main(String[] args){ // Create binary tree shown in above figure Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.right.left.right = newNode(8); root.right.right.right = newNode(9); Console.WriteLine(largestUinquePath(root));}}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to find count of distinct nodes // on a path with maximum distinct nodes. class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // A utility function to create a new Binary // Tree node function newNode(data) { let temp = new Node(data); return temp; } function largestUinquePathUtil(node, m) { if (node == null) return m.size; // put this node into hash if(m.has(node.data)) { m.set(node.data, m.get(node.data) + 1); } else { m.set(node.data, 1); } let max_path = Math.max(largestUinquePathUtil(node.left, m), largestUinquePathUtil(node.right, m)); // remove current node from path "hash" if(m.has(node.data)) { m.set(node.data, m.get(node.data) - 1); } // if we reached a condition where all duplicate value // of current node is deleted if (m.get(node.data) == 0) m.delete(node.data); return max_path; } // A utility function to find long unique value path function largestUinquePath(node) { if (node == null) return 0; // hash that store all node value let hash = new Map(); // return max length unique value path return largestUinquePathUtil(node, hash); } // Create binary tree shown in above figure let root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(6); root.right.right = newNode(7); root.right.left.right = newNode(8); root.right.right.right = newNode(9); document.write(largestUinquePath(root)); </script> |
Output:
4
Time Complexity :O(n)
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