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Number of children of given node in n-ary Tree
  • Difficulty Level : Hard
  • Last Updated : 02 Nov, 2020
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Given a node x, find the number of children of x(if it exists) in the given n-ary tree.

Example :

Input : x = 50
Output : 3
Explanation : 50 has 3 children having values 40, 100 and 20.

Approach :



  • Initialize the number of children as 0.
  • For every node in the n-ary tree, check if its value is equal to x or not. If yes, then return the number of children.
  • If the value of x is not equal to the current node then, push all the children of current node in the queue.
  • Keep Repeating the above step until the queue becomes empty.

Below is the implementation of the above idea :

C++




// C++ program to find number
// of children of given node
#include <bits/stdc++.h>
using namespace std;
  
// Represents a node of an n-ary tree
class Node {
  
public:
    int key;
    vector<Node*> child;
  
    Node(int data)
    {
        key = data;
    }
};
  
// Function to calculate number
// of children of given node
int numberOfChildren(Node* root, int x)
{
    // initialize the numChildren as 0
    int numChildren = 0;
  
    if (root == NULL)
        return 0;
  
    // Creating a queue and pushing the root
    queue<Node*> q;
    q.push(root);
  
    while (!q.empty()) {
        int n = q.size();
  
        // If this node has children
        while (n > 0) {
  
            // Dequeue an item from queue and
            // check if it is equal to x
            // If YES, then return number of children
            Node* p = q.front();
            q.pop();
            if (p->key == x) {
                numChildren = numChildren + p->child.size();
                return numChildren;
            }
  
            // Enqueue all children of the dequeued item
            for (int i = 0; i < p->child.size(); i++)
                q.push(p->child[i]);
            n--;
        }
    }
    return numChildren;
}
  
// Driver program
int main()
{
    // Creating a generic tree
    Node* root = new Node(20);
    (root->child).push_back(new Node(2));
    (root->child).push_back(new Node(34));
    (root->child).push_back(new Node(50));
    (root->child).push_back(new Node(60));
    (root->child).push_back(new Node(70));
    (root->child[0]->child).push_back(new Node(15));
    (root->child[0]->child).push_back(new Node(20));
    (root->child[1]->child).push_back(new Node(30));
    (root->child[2]->child).push_back(new Node(40));
    (root->child[2]->child).push_back(new Node(100));
    (root->child[2]->child).push_back(new Node(20));
    (root->child[0]->child[1]->child).push_back(new Node(25));
    (root->child[0]->child[1]->child).push_back(new Node(50));
  
    // Node whose number of
    // children is to be calculated
    int x = 50;
  
    // Function calling
    cout << numberOfChildren(root, x) << endl;
  
    return 0;
}

Java




// Java program to find number
// of children of given node
import java.util.*;
  
class GFG
{
  
// Represents a node of an n-ary tree
static class Node
{
    int key;
    Vector<Node> child = new Vector<>();
  
    Node(int data)
    {
        key = data;
    }
};
  
// Function to calculate number
// of children of given node
static int numberOfChildren(Node root, int x)
{
    // initialize the numChildren as 0
    int numChildren = 0;
  
    if (root == null)
        return 0;
  
    // Creating a queue and pushing the root
    Queue<Node> q = new LinkedList<Node>();
    q.add(root);
  
    while (!q.isEmpty())
    {
        int n = q.size();
  
        // If this node has children
        while (n > 0
        {
  
            // Dequeue an item from queue and
            // check if it is equal to x
            // If YES, then return number of children
            Node p = q.peek();
            q.remove();
            if (p.key == x) 
            {
                numChildren = numChildren +
                              p.child.size();
                return numChildren;
            }
  
            // Enqueue all children of the dequeued item
            for (int i = 0; i < p.child.size(); i++)
                q.add(p.child.get(i));
            n--;
        }
    }
    return numChildren;
}
  
// Driver Code
public static void main(String[] args) 
{
      
    // Creating a generic tree
    Node root = new Node(20);
    (root.child).add(new Node(2));
    (root.child).add(new Node(34));
    (root.child).add(new Node(50));
    (root.child).add(new Node(60));
    (root.child).add(new Node(70));
    (root.child.get(0).child).add(new Node(15));
    (root.child.get(0).child).add(new Node(20));
    (root.child.get(1).child).add(new Node(30));
    (root.child.get(2).child).add(new Node(40));
    (root.child.get(2).child).add(new Node(100));
    (root.child.get(2).child).add(new Node(20));
    (root.child.get(0).child.get(1).child).add(new Node(25));
    (root.child.get(0).child.get(1).child).add(new Node(50));
  
    // Node whose number of
    // children is to be calculated
    int x = 50;
  
    // Function calling
    System.out.println(numberOfChildren(root, x));
}
}
  
// This code is contributed by 29AjayKumar

Python3




# Python3 program to find number
# of children of given node
  
# Node of a linked list 
class Node: 
    def __init__(self, data = None): 
        self.key = data 
        self.child = []
  
# Function to calculate number
# of children of given node
def numberOfChildren( root, x):
  
    # initialize the numChildren as 0
    numChildren = 0
  
    if (root == None):
        return 0
  
    # Creating a queue and appending the root
    q = []
    q.append(root)
  
    while (len(q) > 0) :
        n = len(q)
  
        # If this node has children
        while (n > 0): 
  
            # Dequeue an item from queue and
            # check if it is equal to x
            # If YES, then return number of children
            p = q[0]
            q.pop(0)
            if (p.key == x) :
                numChildren = numChildren + len(p.child)
                return numChildren
              
            i = 0
              
            # Enqueue all children of the dequeued item
            while ( i < len(p.child)):
                q.append(p.child[i])
                i = i + 1
            n = n - 1
  
    return numChildren
  
# Driver program
  
# Creating a generic tree
root = Node(20)
(root.child).append(Node(2))
(root.child).append(Node(34))
(root.child).append(Node(50))
(root.child).append(Node(60))
(root.child).append(Node(70))
(root.child[0].child).append(Node(15))
(root.child[0].child).append(Node(20))
(root.child[1].child).append(Node(30))
(root.child[2].child).append(Node(40))
(root.child[2].child).append(Node(100))
(root.child[2].child).append(Node(20))
(root.child[0].child[1].child).append(Node(25))
(root.child[0].child[1].child).append(Node(50))
  
# Node whose number of
# children is to be calculated
x = 50
  
# Function calling
print( numberOfChildren(root, x) )
  
# This code is contributed by Arnab Kundu

C#




// C# program to find number
// of children of given node
using System;
using System.Collections.Generic;
      
class GFG
{
  
// Represents a node of an n-ary tree
public class Node
{
    public int key;
    public List<Node> child = new List<Node>();
  
    public Node(int data)
    {
        key = data;
    }
};
  
// Function to calculate number
// of children of given node
static int numberOfChildren(Node root, int x)
{
    // initialize the numChildren as 0
    int numChildren = 0;
  
    if (root == null)
        return 0;
  
    // Creating a queue and pushing the root
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
  
    while (q.Count != 0)
    {
        int n = q.Count;
  
        // If this node has children
        while (n > 0) 
        {
  
            // Dequeue an item from queue and
            // check if it is equal to x
            // If YES, then return number of children
            Node p = q.Peek();
            q.Dequeue();
            if (p.key == x) 
            {
                numChildren = numChildren +
                              p.child.Count;
                return numChildren;
            }
  
            // Enqueue all children of the dequeued item
            for (int i = 0; i < p.child.Count; i++)
                q.Enqueue(p.child[i]);
            n--;
        }
    }
    return numChildren;
}
  
// Driver Code
public static void Main(String[] args) 
{
      
    // Creating a generic tree
    Node root = new Node(20);
    (root.child).Add(new Node(2));
    (root.child).Add(new Node(34));
    (root.child).Add(new Node(50));
    (root.child).Add(new Node(60));
    (root.child).Add(new Node(70));
    (root.child[0].child).Add(new Node(15));
    (root.child[0].child).Add(new Node(20));
    (root.child[1].child).Add(new Node(30));
    (root.child[2].child).Add(new Node(40));
    (root.child[2].child).Add(new Node(100));
    (root.child[2].child).Add(new Node(20));
    (root.child[0].child[1].child).Add(new Node(25));
    (root.child[0].child[1].child).Add(new Node(50));
  
    // Node whose number of
    // children is to be calculated
    int x = 50;
  
    // Function calling
    Console.WriteLine(numberOfChildren(root, x));
}
}
  
// This code is contributed by 29AjayKumar
Output:
3

Time Complexity : O(N), where N is the number of nodes in tree.
Auxiliary Space : O(N), where N is the number of nodes in tree.

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