Number of nodes greater than a given value in n-ary tree
Given a n-ary tree and a number x, find and return the number of nodes which are greater than x.
Example:
In the given tree, x = 7
Number of nodes greater than x are 4.
Approach :
The idea is maintain a count variable initialize to 0. Traverse the tree and compare root data with x. If root data is greater than x, increment the count variable and recursively call for all its children.
Below is the implementation of idea.
C++
// C++ program to find number of nodes // greater than x #include <bits/stdc++.h> using namespace std; // Structure of a node of n-ary tree struct Node { int key; vector<Node*> child; }; // Utility function to create // a new tree node Node* newNode(int key) { Node* temp = new Node; temp->key = key; return temp; } // Function to find nuber of nodes // gretaer than x int nodesGreaterThanX(Node* root, int x) { if (root == NULL) return 0; int count = 0; // if current root is greater // than x increment count if (root->key > x) count++; // Number of children of root int numChildren = root->child.size(); // recursively calling for every child for (int i = 0; i < numChildren; i++) { Node* child = root->child[i]; count += nodesGreaterThanX(child, x); } // return the count return count; } // Driver program int main() { /* Let us create below tree * 5 * / | \ * 1 2 3 * / / \ \ * 15 4 5 6 */ Node* root = newNode(5); (root->child).push_back(newNode(1)); (root->child).push_back(newNode(2)); (root->child).push_back(newNode(3)); (root->child[0]->child).push_back(newNode(15)); (root->child[1]->child).push_back(newNode(4)); (root->child[1]->child).push_back(newNode(5)); (root->child[2]->child).push_back(newNode(6)); int x = 5; cout << "Number of nodes greater than " << x << " are "; cout << nodesGreaterThanX(root, x) << endl; return 0; } |
Java
// Java program to find number of nodes // greater than x import java.util.*; // Class representing a Node of an N-ary tree class Node{ int key; ArrayList<Node> child; // Constructor to create a Node Node(int val) { key = val; child = new ArrayList<>(); } } class GFG{ // Recursive function to find number // of nodes greater than x public static int nodesGreaterThanX(Node root, int x) { if (root == null) return 0; int count = 0; // If current root is greater // than x increment count if (root.key > x) count++; // Recursively calling for every // child of current root for(Node child : root.child) { count += nodesGreaterThanX(child, x); } // Return the count return count; } // Driver code public static void main(String[] args) { /* Let us create below tree 5 / | \ 1 2 3 / / \ \ 15 4 5 6 */ Node root = new Node(5); root.child.add(new Node(1)); root.child.add(new Node(2)); root.child.add(new Node(3)); root.child.get(0).child.add(new Node(15)); root.child.get(1).child.add(new Node(4)); root.child.get(1).child.add(new Node(5)); root.child.get(2).child.add(new Node(6)); int x = 5; System.out.print("Number of nodes greater than " + x + " are "); System.out.println(nodesGreaterThanX(root, x)); } } // This code is contributed by jrishabh99 |
Python3
# Python3 program to find number of nodes # greater than x # Structure of a node of n-ary tree class Node: def __init__(self, data): self.key = data self.child = [] # Function to find nuber of nodes # gretaer than x def nodesGreaterThanX(root: Node, x: int) -> int: if root is None: return 0 count = 0 # if current root is greater # than x increment count if root.key > x: count += 1 # Number of children of root numChildren = len(root.child) # recursively calling for every child for i in range(numChildren): child = root.child[i] count += nodesGreaterThanX(child, x) # return the count return count # Driver Code if __name__ == "__main__": ans = 0 k = 25 # Let us create below tree # 5 # / | \ # 1 2 3 # / / \ \ # 15 4 5 6 root = Node(5) (root.child).append(Node(1)) (root.child).append(Node(2)) (root.child).append(Node(3)) (root.child[0].child).append(Node(15)) (root.child[1].child).append(Node(4)) (root.child[1].child).append(Node(5)) (root.child[2].child).append(Node(6)) x = 5 print("Number of nodes greater than % d are % d" % (x, nodesGreaterThanX(root, x))) # This code is contributed by # sanjeev2552 |
C#
// C# program to find number of nodes // greater than x using System;using System.Collections.Generic; // Class representing a Node of an N-ary tree public class Node{ public int key; public List<Node> child; // Constructor to create a Node public Node(int val) { key = val; child = new List<Node>(); } } class GFG{ // Recursive function to find number // of nodes greater than x public static int nodesGreaterThanX(Node root, int x) { if (root == null) return 0; int count = 0; // If current root is greater // than x increment count if (root.key > x) count++; // Recursively calling for every // child of current root foreach(Node child in root.child) { count += nodesGreaterThanX(child, x); } // Return the count return count; } // Driver code public static void Main(String[] args) { /* Let us create below tree 5 / | \ 1 2 3 / / \ \ 15 4 5 6 */ Node root = new Node(5); root.child.Add(new Node(1)); root.child.Add(new Node(2)); root.child.Add(new Node(3)); root.child[0].child.Add(new Node(15)); root.child[1].child.Add(new Node(4)); root.child[1].child.Add(new Node(5)); root.child[2].child.Add(new Node(6)); int x = 5; Console.Write("Number of nodes greater than " + x + " are "); Console.WriteLine(nodesGreaterThanX(root, x)); } } // This code is contributed by Amit Katiyar |
Output:
Number of nodes greater than 5 are 2
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