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

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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

                    

Javascript

// JavaScript program to find number of nodes
// greater than x
 
// Structure of a node of n-ary tree
class Node {
    constructor(key) {
        this.key = key;
        this.child = [];
    }
}
 
// Utility function to create
// a new tree node
function newNode(key) {
    return new Node(key);
}
 
// Function to find number of nodes
// greater than x
function nodesGreaterThanX(root, x) {
    if (root == null)
        return 0;
 
    let count = 0;
 
    // if current root is greater
    // than x increment count
    if (root.key > x)
        count++;
 
    // Number of children of root
    const numChildren = root.child.length;
 
    // recursively calling for every child
    for (let i = 0; i < numChildren; i++) {
        const child = root.child[i];
        count += nodesGreaterThanX(child, x);
    }
 
    // return the count
    return count;
}
 
// Driver program
 
    /* Let us create below tree
    *         5
    *         / | \
    *     1 2 3
    *     / / \ \
    *     15 4 5 6
    */
 
    const root = newNode(5);
    root.child.push(newNode(1));
    root.child.push(newNode(2));
    root.child.push(newNode(3));
    root.child[0].child.push(newNode(15));
    root.child[1].child.push(newNode(4));
    root.child[1].child.push(newNode(5));
    root.child[2].child.push(newNode(6));
 
    const x = 5;
 
    console.log("Number of nodes greater than " + x + " are " + nodesGreaterThanX(root, x));

                    

Output
Number of nodes greater than 5 are 2

Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(h) where h is the height of the binary tree.

 



Last Updated : 13 Mar, 2023
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