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Number of nodes greater than a given value in n-ary tree
  • Difficulty Level : Easy
  • Last Updated : 04 Nov, 2020

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

tree

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 
            
        / | \ 
        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 
          
        / | \ 
       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

This article is contributed by Chhavi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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