Skip to content
Related Articles

Related Articles

Number of nodes greater than a given value in n-ary tree

View Discussion
Improve Article
Save Article
Like Article
  • Difficulty Level : Easy
  • Last Updated : 04 Aug, 2021

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 number of nodes
// greater 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 number of nodes
# greater 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

This article is contributed by Chhavi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!