Number of siblings of a given Node in n-ary Tree

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++ 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 {
    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;
    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();
        // 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;
// 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;




Time Complexity : O(N^2), where N is the number of nodes in tree.
Auxiliary Space : O(N), where N is the number of nodes in tree.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :


Please write to us at to report any issue with the above content.