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Number of siblings of a given Node in n-ary Tree
  • Difficulty Level : Easy
  • Last Updated : 02 Nov, 2020

Given an N-ary tree, find the number of siblings of given node x. Assume that x exists in the given n-ary tree.

Example :

Input : 30
Output : 3

Approach : For every node in the given n-ary tree, push the children of the current node in the queue. While adding the children of current node in queue, check if any children is equal to the given value x or not. If yes, then return the number of siblings of x.



Below is the implementation of the above idea :

C++




// C++ program to find number
// of siblings of a 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 siblings of a given node
int numberOfSiblings(Node* root, int x)
{
    if (root == NULL)
        return 0;
  
    // Creating a queue and
    // pushing the root
    queue<Node*> q;
    q.push(root);
  
    while (!q.empty()) {
        // 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();
  
        // Enqueue all children of
        // the dequeued item
        for (int i = 0; i < p->child.size(); i++) {
            // If the value of children
            // is equal to x, then return
            // the number of siblings
            if (p->child[i]->key == x)
                return p->child.size() - 1;
            q.push(p->child[i]);
        }
    }
}
  
// Driver program
int main()
{
    // Creating a generic tree as shown in above figure
    Node* root = new Node(50);
    (root->child).push_back(new Node(2));
    (root->child).push_back(new Node(30));
    (root->child).push_back(new Node(14));
    (root->child).push_back(new Node(60));
    (root->child[0]->child).push_back(new Node(15));
    (root->child[0]->child).push_back(new Node(25));
    (root->child[0]->child[1]->child).push_back(new Node(70));
    (root->child[0]->child[1]->child).push_back(new Node(100));
    (root->child[1]->child).push_back(new Node(6));
    (root->child[1]->child).push_back(new Node(1));
    (root->child[2]->child).push_back(new Node(7));
    (root->child[2]->child[0]->child).push_back(new Node(17));
    (root->child[2]->child[0]->child).push_back(new Node(99));
    (root->child[2]->child[0]->child).push_back(new Node(27));
    (root->child[3]->child).push_back(new Node(16));
  
    // Node whose number of
    // siblings is to be calculated
    int x = 100;
  
    // Function calling
    cout << numberOfSiblings(root, x) << endl;
  
    return 0;
}

Java




// Java program to find number 
// of siblings of a given node 
import java.util.*;
  
class GFG
{
      
// Represents a node of an n-ary tree 
static class Node 
    int key; 
    Vector<Node> child; 
  
    Node(int data) 
    
        key = data; 
        child = new Vector<Node>();
    
}; 
  
// Function to calculate number 
// of siblings of a given node 
static int numberOfSiblings(Node root, int x) 
    if (root == null
        return 0
  
    // Creating a queue and 
    // pushing the root 
    Queue<Node> q = new LinkedList<>(); 
    q.add(root); 
  
    while (q.size() > 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(); 
  
        // Enqueue all children of 
        // the dequeued item 
        for (int i = 0; i < p.child.size(); i++) 
        
            // If the value of children 
            // is equal to x, then return 
            // the number of siblings 
            if (p.child.get(i).key == x) 
                return p.child.size() - 1
            q.add(p.child.get(i)); 
        
    
    return -1;
  
// Driver code
public static void main(String args[])
    // Creating a generic tree as shown in above figure 
    Node root = new Node(50); 
    (root.child).add(new Node(2)); 
    (root.child).add(new Node(30)); 
    (root.child).add(new Node(14)); 
    (root.child).add(new Node(60)); 
    (root.child.get(0).child).add(new Node(15)); 
    (root.child.get(0).child).add(new Node(25)); 
    (root.child.get(0).child.get(1).child).add(new Node(70)); 
    (root.child.get(0).child.get(1).child).add(new Node(100)); 
    (root.child.get(1).child).add(new Node(6)); 
    (root.child.get(1).child).add(new Node(1)); 
    (root.child.get(2).child).add(new Node(7)); 
    (root.child.get(2).child.get(0).child).add(new Node(17)); 
    (root.child.get(2).child.get(0).child).add(new Node(99)); 
    (root.child.get(2).child.get(0).child).add(new Node(27)); 
    (root.child.get(3).child).add(new Node(16)); 
  
    // Node whose number of 
    // siblings is to be calculated 
    int x = 100
  
    // Function calling 
    System.out.println( numberOfSiblings(root, x) ); 
}
  
// This code is contributed by Arnab Kundu

Python3




# Python3 program to find number 
# of siblings of a given node 
from queue import Queue 
  
# Represents a node of an n-ary tree 
class newNode: 
    def __init__(self,data):
        self.child = []
        self.key = data
  
# Function to calculate number 
# of siblings of a given node 
def numberOfSiblings(root, x):
    if (root == None): 
        return 0
  
    # Creating a queue and 
    # pushing the root 
    q = Queue()
    q.put(root) 
  
    while (not q.empty()): 
          
        # Dequeue an item from queue and 
        # check if it is equal to x If YES, 
        # then return number of children 
        p = q.queue[0
        q.get() 
  
        # Enqueue all children of 
        # the dequeued item
        for i in range(len(p.child)):
              
            # If the value of children 
            # is equal to x, then return 
            # the number of siblings 
            if (p.child[i].key == x): 
                return len(p.child) - 1
            q.put(p.child[i])
  
# Driver Code
if __name__ == '__main__':
  
    # Creating a generic tree as 
    # shown in above figure 
    root = newNode(50
    (root.child).append(newNode(2)) 
    (root.child).append(newNode(30)) 
    (root.child).append(newNode(14)) 
    (root.child).append(newNode(60)) 
    (root.child[0].child).append(newNode(15)) 
    (root.child[0].child).append(newNode(25)) 
    (root.child[0].child[1].child).append(newNode(70)) 
    (root.child[0].child[1].child).append(newNode(100)) 
    (root.child[1].child).append(newNode(6)) 
    (root.child[1].child).append(newNode(1)) 
    (root.child[2].child).append(newNode(7)) 
    (root.child[2].child[0].child).append(newNode(17)) 
    (root.child[2].child[0].child).append(newNode(99)) 
    (root.child[2].child[0].child).append(newNode(27)) 
    (root.child[3].child).append(newNode(16)) 
  
    # Node whose number of 
    # siblings is to be calculated 
    x = 100
  
    # Function calling 
    print(numberOfSiblings(root, x))
  
# This code is contributed by PranchalK

C#




// C# program to find number 
// of siblings of a 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; 
  
    public Node(int data) 
    
        key = data; 
        child = new List<Node>();
    
}; 
  
// Function to calculate number 
// of siblings of a given node 
static int numberOfSiblings(Node root, int x) 
    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) 
    
        // 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(); 
  
        // Enqueue all children of 
        // the dequeued item 
        for (int i = 0; i < p.child.Count; i++) 
        
            // If the value of children 
            // is equal to x, then return 
            // the number of siblings 
            if (p.child[i].key == x) 
                return p.child.Count - 1; 
            q.Enqueue(p.child[i]); 
        
    
    return -1;
  
// Driver code
public static void Main(String []args)
    // Creating a generic tree
    // as shown in above figure 
    Node root = new Node(50); 
    (root.child).Add(new Node(2)); 
    (root.child).Add(new Node(30)); 
    (root.child).Add(new Node(14)); 
    (root.child).Add(new Node(60)); 
    (root.child[0].child).Add(new Node(15)); 
    (root.child[0].child).Add(new Node(25)); 
    (root.child[0].child[1].child).Add(new Node(70)); 
    (root.child[0].child[1].child).Add(new Node(100)); 
    (root.child[1].child).Add(new Node(6)); 
    (root.child[1].child).Add(new Node(1)); 
    (root.child[2].child).Add(new Node(7)); 
    (root.child[2].child[0].child).Add(new Node(17)); 
    (root.child[2].child[0].child).Add(new Node(99)); 
    (root.child[2].child[0].child).Add(new Node(27)); 
    (root.child[3].child).Add(new Node(16)); 
  
    // Node whose number of 
    // siblings is to be calculated 
    int x = 100; 
  
    // Function calling 
    Console.WriteLine( numberOfSiblings(root, x)); 
}
}
  
// This code is contributed by PrinciRaj1992 
Output:
1

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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