# Number of children of given node in n-ary Tree

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 ` `using` `namespace` `std;`   `// Represents a node of an n-ary tree` `class` `Node {`   `public``:` `    ``int` `key;` `    ``vector 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 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 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 q = ``new` `LinkedList();` `    ``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 child = ``new` `List();`   `    ``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 q = ``new` `Queue();` `    ``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`

## Javascript

 ``

Output:

`3`

Complexity Analysis:

• 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.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next