# Count of nodes in a Binary tree with immediate children as its factors

Last Updated : 05 Mar, 2023

Given a Binary Tree, the task is to print the count of nodes whose immediate children are its factors

Examples:

```Input:
1
/   \
15     20
/  \   /  \
3    5 4     2
\    /
2  3
Output: 2
Explanation:
Children of 15 (3, 5)
are factors of 15
Children of 20 (4, 2)
are factors of 20

Input:
7
/  \
210   14
/  \      \
70   14     30
/ \         / \
2   5       10  15
/
23
Output:3
Explanation:
Children of 210 (70, 14)
are factors of 210
Children of 70 (2, 5)
are factors of 70
Children of 30 (10, 15)
are factors of 30```

Approach: In order to solve this problem, we need to traverse the given Binary Tree in Level Order fashion and for every node with both children, check if both the children have values which are factors of the value of the current node. If true, then count such nodes and print it at the end.

Below is the implementation of the above approach:

## C++

 `// C++ program for Counting nodes` `// whose immediate children` `// are its factors`   `#include ` `using` `namespace` `std;`   `// A Tree node` `struct` `Node {` `    ``int` `key;` `    ``struct` `Node *left, *right;` `};`   `// Utility function to create a new node` `Node* newNode(``int` `key)` `{` `    ``Node* temp = ``new` `Node;` `    ``temp->key = key;` `    ``temp->left = temp->right = NULL;` `    ``return` `(temp);` `}`   `// Function to check and print if` `// immediate children of a node` `// are its factors or not` `bool` `areChilrenFactors(` `    ``struct` `Node* parent,` `    ``struct` `Node* a,` `    ``struct` `Node* b)` `{` `    ``if` `(parent->key % a->key == 0` `        ``&& parent->key % b->key == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Function to get the` `// count of full Nodes in` `// a binary tree` `unsigned ``int` `getCount(``struct` `Node* node)` `{` `    ``// If tree is empty` `    ``if` `(!node)` `        ``return` `0;` `    ``queue q;`   `    ``// Do level order traversal` `    ``// starting from root` `    ``int` `count = 0;` `    ``// Store the number of nodes` `    ``// with both children as factors` `    ``q.push(node);` `    ``while` `(!q.empty()) {` `        ``struct` `Node* temp = q.front();` `        ``q.pop();`   `        ``if` `(temp->left && temp->right) {` `            ``if` `(areChilrenFactors(` `                    ``temp, temp->left,` `                    ``temp->right))` `                ``count++;` `        ``}`   `        ``if` `(temp->left != NULL)` `            ``q.push(temp->left);` `        ``if` `(temp->right != NULL)` `            ``q.push(temp->right);` `    ``}` `    ``return` `count;` `}`   `// Function to find total no of nodes` `// In a given binary tree` `int` `findSize(``struct` `Node* node)` `{` `    ``// Base condition` `    ``if` `(node == NULL)` `        ``return` `0;`   `    ``return` `1` `           ``+ findSize(node->left)` `           ``+ findSize(node->right);` `}`   `// Driver Code` `int` `main()` `{` `    ``/*        10 ` `            ``/ \ ` `           ``40 36 ` `              ``/ \ ` `             ``18  12 ` `             ``/ \ / \ ` `            ``2  6 3 4 ` `                  ``/ ` `                 ``7` `    ``*/`   `    ``// Create Binary Tree as shown` `    ``Node* root = newNode(10);`   `    ``root->left = newNode(40);` `    ``root->right = newNode(36);`   `    ``root->right->left = newNode(18);` `    ``root->right->right = newNode(12);`   `    ``root->right->left->left = newNode(2);` `    ``root->right->left->right = newNode(6);` `    ``root->right->right->left = newNode(3);` `    ``root->right->right->right = newNode(4);` `    ``root->right->right->right->left = newNode(7);`   `    ``// Print all nodes having` `    ``// children as their factors` `    ``cout << getCount(root) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java program for Counting nodes` `// whose immediate children` `// are its factors` ` ``import` `java.util.*;`   `class` `GFG{` ` `  `// A Tree node` `static` `class` `Node {` `    ``int` `key;` `    ``Node left, right;` `};` ` `  `// Utility function to create a new node` `static` `Node newNode(``int` `key)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.key = key;` `    ``temp.left = temp.right = ``null``;` `    ``return` `(temp);` `}` ` `  `// Function to check and print if` `// immediate children of a node` `// are its factors or not` `static` `boolean` `areChilrenFactors(` `    ``Node parent,` `    ``Node a,` `    ``Node b)` `{` `    ``if` `(parent.key % a.key == ``0` `        ``&& parent.key % b.key == ``0``)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}` ` `  `// Function to get the` `// count of full Nodes in` `// a binary tree` `static` `int` `getCount(Node node)` `{` `    ``// If tree is empty` `    ``if` `(node==``null``)` `        ``return` `0``;` `    ``Queue q = ``new` `LinkedList();` ` `  `    ``// Do level order traversal` `    ``// starting from root` `    ``int` `count = ``0``;`   `    ``// Store the number of nodes` `    ``// with both children as factors` `    ``q.add(node);` `    ``while` `(!q.isEmpty()) {` `        ``Node temp = q.peek();` `        ``q.remove();` ` `  `        ``if` `(temp.left!=``null` `&& temp.right!=``null``) {` `            ``if` `(areChilrenFactors(` `                    ``temp, temp.left,` `                    ``temp.right))` `                ``count++;` `        ``}` ` `  `        ``if` `(temp.left != ``null``)` `            ``q.add(temp.left);` `        ``if` `(temp.right != ``null``)` `            ``q.add(temp.right);` `    ``}` `    ``return` `count;` `}` ` `  `// Function to find total no of nodes` `// In a given binary tree` `static` `int` `findSize(Node node)` `{` `    ``// Base condition` `    ``if` `(node == ``null``)` `        ``return` `0``;` ` `  `    ``return` `1` `           ``+ findSize(node.left)` `           ``+ findSize(node.right);` `}` ` `  `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``/*        10 ` `            ``/ \ ` `           ``40 36 ` `              ``/ \ ` `             ``18  12 ` `             ``/ \ / \ ` `            ``2  6 3 4 ` `                  ``/ ` `                 ``7` `    ``*/` ` `  `    ``// Create Binary Tree as shown` `    ``Node root = newNode(``10``);` ` `  `    ``root.left = newNode(``40``);` `    ``root.right = newNode(``36``);` ` `  `    ``root.right.left = newNode(``18``);` `    ``root.right.right = newNode(``12``);` ` `  `    ``root.right.left.left = newNode(``2``);` `    ``root.right.left.right = newNode(``6``);` `    ``root.right.right.left = newNode(``3``);` `    ``root.right.right.right = newNode(``4``);` `    ``root.right.right.right.left = newNode(``7``);` ` `  `    ``// Print all nodes having` `    ``// children as their factors` `    ``System.out.print(getCount(root) +``"\n"``);` ` `  `}` `}`   `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program for counting nodes` `# whose immediate children` `# are its factors` `from` `collections ``import` `deque as queue`   `# A Binary Tree Node` `class` `Node:` `    `  `    ``def` `__init__(``self``, key):` `        `  `        ``self``.data ``=` `key` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to check and print if` `# immediate children of a node` `# are its factors or not` `def` `areChildrenFactors(parent, a, b):` `    `  `    ``if` `(parent.data ``%` `a.data ``=``=` `0` `and` `        ``parent.data ``%` `b.data ``=``=` `0``):` `        ``return` `True` `    ``else``:` `        ``return` `False`   `# Function to get the` `# count of full Nodes in` `# a binary tree` `def` `getCount(node):` `    `  `    ``# Base Case` `    ``if` `(``not` `node):` `        ``return` `0`   `    ``q ``=` `queue()`   `    ``# Do level order traversal` `    ``# starting from root` `    ``count ``=` `0` `    `  `    ``# Store the number of nodes` `    ``# with both children as factors` `    ``q.append(node)`   `    ``while` `(``len``(q) > ``0``):` `        ``temp ``=` `q.popleft()` `        ``#q.pop()`   `        ``if` `(temp.left ``and` `temp.right):` `            ``if` `(areChildrenFactors(temp, temp.left,` `                                         ``temp.right)):` `                ``count ``+``=` `1`   `        ``if` `(temp.left !``=` `None``):` `            ``q.append(temp.left)` `        ``if` `(temp.right !``=` `None``):` `            ``q.append(temp.right)`   `    ``return` `count`   `# Function to find total` `# number of nodes` `# In a given binary tree` `def` `findSize(node):` `    `  `    ``# Base condition` `    ``if` `(node ``=``=` `None``):` `        ``return` `0`   `    ``return` `(``1` `+` `findSize(node.left) ``+`  `                ``findSize(node.right))`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# /*        10` `    ``#          / \` `    ``#         40  36` `    ``#            /  \` `    ``#           18   12` `    ``#           / \  / \` `    ``#          2   6 3  4` `    ``#                  /` `    ``#                 7` `    ``#  */`   `    ``# Create Binary Tree` `    ``root ``=` `Node(``10``)` `    ``root.left ``=` `Node(``40``)` `    ``root.right ``=` `Node(``36``)`   `    ``root.right.left ``=` `Node(``18``)` `    ``root.right.right ``=` `Node(``12``)`   `    ``root.right.left.left ``=` `Node(``2``)` `    ``root.right.left.right ``=` `Node(``6``)` `    ``root.right.right.left ``=` `Node(``3``)` `    ``root.right.right.right ``=` `Node(``4``)` `    ``root.right.right.right.left ``=` `Node(``7``)`   `    ``# Print all nodes having` `    ``# children as their factors` `    ``print``(getCount(root))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for Counting nodes` `// whose immediate children` `// are its factors` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{` `  `  `// A Tree node` `class` `Node {` `    ``public` `int` `key;` `    ``public` `Node left, right;` `};` `  `  `// Utility function to create a new node` `static` `Node newNode(``int` `key)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.key = key;` `    ``temp.left = temp.right = ``null``;` `    ``return` `(temp);` `}` `  `  `// Function to check and print if` `// immediate children of a node` `// are its factors or not` `static` `bool` `areChilrenFactors(` `    ``Node parent,` `    ``Node a,` `    ``Node b)` `{` `    ``if` `(parent.key % a.key == 0` `        ``&& parent.key % b.key == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}` `  `  `// Function to get the` `// count of full Nodes in` `// a binary tree` `static` `int` `getCount(Node node)` `{` `    ``// If tree is empty` `    ``if` `(node == ``null``)` `        ``return` `0;` `    ``List q = ``new` `List();` `  `  `    ``// Do level order traversal` `    ``// starting from root` `    ``int` `count = 0;` ` `  `    ``// Store the number of nodes` `    ``// with both children as factors` `    ``q.Add(node);` `    ``while` `(q.Count != 0) {` `        ``Node temp = q[0];` `        ``q.RemoveAt(0);` `  `  `        ``if` `(temp.left!=``null` `&& temp.right != ``null``) {` `            ``if` `(areChilrenFactors(` `                    ``temp, temp.left,` `                    ``temp.right))` `                ``count++;` `        ``}` `  `  `        ``if` `(temp.left != ``null``)` `            ``q.Add(temp.left);` `        ``if` `(temp.right != ``null``)` `            ``q.Add(temp.right);` `    ``}` `    ``return` `count;` `}` `  `  `// Function to find total no of nodes` `// In a given binary tree` `static` `int` `findSize(Node node)` `{` `    ``// Base condition` `    ``if` `(node == ``null``)` `        ``return` `0;` `  `  `    ``return` `1` `           ``+ findSize(node.left)` `           ``+ findSize(node.right);` `}` `  `  `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``/*        10 ` `            ``/ \ ` `           ``40 36 ` `              ``/ \ ` `             ``18  12 ` `             ``/ \ / \ ` `            ``2  6 3 4 ` `                  ``/ ` `                 ``7` `    ``*/` `  `  `    ``// Create Binary Tree as shown` `    ``Node root = newNode(10);` `  `  `    ``root.left = newNode(40);` `    ``root.right = newNode(36);` `  `  `    ``root.right.left = newNode(18);` `    ``root.right.right = newNode(12);` `  `  `    ``root.right.left.left = newNode(2);` `    ``root.right.left.right = newNode(6);` `    ``root.right.right.left = newNode(3);` `    ``root.right.right.right = newNode(4);` `    ``root.right.right.right.left = newNode(7);` `  `  `    ``// Print all nodes having` `    ``// children as their factors` `    ``Console.Write(getCount(root) +``"\n"``);` `  `  `}` `}`   `// This code is contributed by sapnasingh4991`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N) where N is the number of nodes in the binary tree.
Auxiliary Space: O(N), as Level order uses N space for the call stack, where N is the number of nodes in the binary tree.

Efficient Method 2 (Using only one traversal without extra space):

we simply traverse the binary tree in preorder fashion and if a node has left and right child then we check that these childs are the factors of root node or not. If yes we increment the count variable.

Below is the implementation of above approach:

## C++

 `// C++ program for Counting nodes whose` `// immediate children are its factors` `#include ` `using` `namespace` `std;`   `// A Tree node` `struct` `Node {` `    ``int` `data;` `    ``struct` `Node* left;` `    ``struct` `Node* right;` `};`   `// Utility function to create a new node` `Node* newNode(``int` `key)` `{` `    ``Node* temp = ``new` `Node;` `    ``temp->data = key;` `    ``temp->left = temp->right = NULL;` `    ``return` `temp;` `}`   `// Function to check and print if immediate` `// children of a node are its factors or not` `bool` `areChilrenFactors(Node* parent, Node* a, Node* b)` `{` `    ``if` `(parent->data % a->data == 0 && parent->data % b->data == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Function to get the count of full Nodes` `// in a binary tree` `void` `getCount(Node* root, ``int``& count)` `{` `    ``if` `(root == NULL)` `        ``return``;` `    ``if` `(root->left != NULL && root->right != NULL) {` `        ``if` `(areChilrenFactors(root, root->left, root->right)) {` `            ``count++;` `        ``}` `    ``}` `    ``getCount(root->left, count);` `    ``getCount(root->right, count);` `}`   `// Driver Code` `int` `main()` `{` `    ``/*     10` `            ``/ \` `       ``40 36` `          ``/ \` `             ``18 12` `        ``/ \ / \` `            ``2 6 3 4` `                      ``/` `                     ``7` `    ``*/`   `    ``// Create Binary Tree as shown` `    ``Node* root = newNode(10);`   `    ``root->left = newNode(40);` `    ``root->right = newNode(36);`   `    ``root->right->left = newNode(18);` `    ``root->right->right = newNode(12);`   `    ``root->right->left->left = newNode(2);` `    ``root->right->left->right = newNode(6);` `    ``root->right->right->left = newNode(3);` `    ``root->right->right->right = newNode(4);` `    ``root->right->right->right->left = newNode(7);`   `    ``// Print all nodes having` `    ``// children as their factors` `    ``int` `count = 0;` `    ``getCount(root, count);` `    ``cout << count << endl;` `    ``return` `0;` `}`   `// This code is contributed by Yash Agarwal(yashagarwal2852002)`

## Java

 `// Java program for Counting nodes whose` `// immediate children are its factors`   `import` `java.io.*;`   `// A Tree node` `class` `Node {` `    ``int` `data;` `    ``Node left, right;` `    ``Node(``int` `key) {` `        ``data = key;` `        ``left = right = ``null``;` `    ``}` `}`   `// Main class` `class` `Main {` `    ``// Function to check and print if immediate` `    ``// children of a node are its factors or not` `    ``static` `boolean` `areChildrenFactors(Node parent, Node a, Node b) {` `        ``if` `(parent.data % a.data == ``0` `&& parent.data % b.data == ``0``)` `            ``return` `true``;` `        ``else` `            ``return` `false``;` `    ``}`   `    ``// Function to get the count of full Nodes` `    ``// in a binary tree` `    ``static` `void` `getCount(Node root, ``int``[] count) {` `        ``if` `(root == ``null``)` `            ``return``;` `        ``if` `(root.left != ``null` `&& root.right != ``null``) {` `            ``if` `(areChildrenFactors(root, root.left, root.right)) {` `                ``count[``0``]++;` `            ``}` `        ``}` `        ``getCount(root.left, count);` `        ``getCount(root.right, count);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args) {` `        ``/*        10` `                  ``/ \` `                ``40  36` `                ``/ \` `               ``18  12` `              ``/ \  / \` `             ``2  6  3  4` `                   ``/` `                  ``7` `        ``*/`   `        ``// Create Binary Tree as shown` `        ``Node root = ``new` `Node(``10``);`   `        ``root.left = ``new` `Node(``40``);` `        ``root.right = ``new` `Node(``36``);`   `        ``root.right.left = ``new` `Node(``18``);` `        ``root.right.right = ``new` `Node(``12``);`   `        ``root.right.left.left = ``new` `Node(``2``);` `        ``root.right.left.right = ``new` `Node(``6``);` `        ``root.right.right.left = ``new` `Node(``3``);` `        ``root.right.right.right = ``new` `Node(``4``);` `        ``root.right.right.right.left = ``new` `Node(``7``);`   `        ``// Print all nodes having` `        ``// children as their factors` `        ``int``[] count = {``0``};` `        ``getCount(root, count);` `        ``System.out.println(count[``0``]);` `    ``}` `}`

## Python3

 `# Python program for counting nodes whose` `# immediate children are its factors`   `# A Tree node` `class` `Node:` `    ``def` `__init__(``self``, key):` `        ``self``.data ``=` `key` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None`   `# Function to check and print if immediate` `# children of a node are its factors or not` `def` `areChildrenFactors(parent, a, b):` `    ``if` `parent.data ``%` `a.data ``=``=` `0` `and` `parent.data ``%` `b.data ``=``=` `0``:` `        ``return` `True` `    ``else``:` `        ``return` `False`   `# Function to get the count of full Nodes` `# in a binary tree` `def` `getCount(root, count):` `    ``if` `root ``is` `None``:` `        ``return` `    ``if` `root.left ``is` `not` `None` `and` `root.right ``is` `not` `None``:` `        ``if` `areChildrenFactors(root, root.left, root.right):` `            ``count[``0``] ``+``=` `1` `    ``getCount(root.left, count)` `    ``getCount(root.right, count)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``# Create Binary Tree as shown` `    ``root ``=` `Node(``10``)`   `    ``root.left ``=` `Node(``40``)` `    ``root.right ``=` `Node(``36``)`   `    ``root.right.left ``=` `Node(``18``)` `    ``root.right.right ``=` `Node(``12``)`   `    ``root.right.left.left ``=` `Node(``2``)` `    ``root.right.left.right ``=` `Node(``6``)` `    ``root.right.right.left ``=` `Node(``3``)` `    ``root.right.right.right ``=` `Node(``4``)` `    ``root.right.right.right.left ``=` `Node(``7``)`   `    ``# Print all nodes having children` `    ``# as their factors` `    ``count ``=` `[``0``]` `    ``getCount(root, count)` `    ``print``(count[``0``])`

## C#

 `using` `System;`   `// A Tree node` `class` `Node` `{` `  ``public` `int` `data;` `  ``public` `Node left, right;`   `  ``public` `Node(``int` `key)` `  ``{` `    ``data = key;` `    ``left = right = ``null``;` `  ``}` `}`   `// Main class` `class` `MainClass` `{` `  ``// Function to check and print if immediate` `  ``// children of a node are its factors or not` `  ``static` `bool` `AreChildrenFactors(Node parent, Node a, Node b)` `  ``{` `    ``if` `(parent.data % a.data == 0 && parent.data % b.data == 0)` `      ``return` `true``;` `    ``else` `      ``return` `false``;` `  ``}`   `  ``// Function to get the count of full Nodes` `  ``// in a binary tree` `  ``static` `void` `GetCount(Node root, ``int``[] count)` `  ``{` `    ``if` `(root == ``null``)` `      ``return``;` `    ``if` `(root.left != ``null` `&& root.right != ``null``)` `    ``{` `      ``if` `(AreChildrenFactors(root, root.left, root.right))` `      ``{` `        ``count[0]++;` `      ``}` `    ``}` `    ``GetCount(root.left, count);` `    ``GetCount(root.right, count);` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main()` `  ``{` `    ``/*        10` `                  ``/ \` `                ``40  36` `                ``/ \` `               ``18  12` `              ``/ \  / \` `             ``2  6  3  4` `                   ``/` `                  ``7` `        ``*/`   `    ``// Create Binary Tree as shown` `    ``Node root = ``new` `Node(10);`   `    ``root.left = ``new` `Node(40);` `    ``root.right = ``new` `Node(36);`   `    ``root.right.left = ``new` `Node(18);` `    ``root.right.right = ``new` `Node(12);`   `    ``root.right.left.left = ``new` `Node(2);` `    ``root.right.left.right = ``new` `Node(6);` `    ``root.right.right.left = ``new` `Node(3);` `    ``root.right.right.right = ``new` `Node(4);` `    ``root.right.right.right.left = ``new` `Node(7);`   `    ``// Print all nodes having` `    ``// children as their factors` `    ``int``[] count = {0};` `    ``GetCount(root, count);` `    ``Console.WriteLine(count[0]);` `  ``}` `}`

## Javascript

 `// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)` `// JavaScript program for counting nodes whose` `// immediate children are its factors` `class Node{` `    ``constructor(data){` `        ``this``.data = data;` `        ``this``.left = ``null``;` `        ``this``.right = ``null``;` `    ``}` `}`   `// utility function to create a new node` `function` `newNode(key){` `    ``return` `new` `Node(key);` `}`   `// Function to check and print if immediate` `// children of a node are its factors or not` `function` `areChilrenFactors(parent, a, b){` `    ``if` `(parent.data % a.data == 0 && parent.data % b.data == 0)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `let count = 0;` `// Function to get the count of full Nodes` `// in a binary tree` `function` `getCount(root)` `{` `    ``if` `(root == ``null``)` `        ``return``;` `    ``if` `(root.left != ``null` `&& root.right != ``null``) {` `        ``if` `(areChilrenFactors(root, root.left, root.right)) {` `            ``count++;` `        ``}` `    ``}` `    ``getCount(root.left);` `    ``getCount(root.right);` `}`   `// Create Binary Tree as shown` `let root = newNode(10);` ` `  `root.left = newNode(40);` `root.right = newNode(36);` ` `  `root.right.left = newNode(18);` `root.right.right = newNode(12);` ` `  `root.right.left.left = newNode(2);` `root.right.left.right = newNode(6);` `root.right.right.left = newNode(3);` `root.right.right.right = newNode(4);` `root.right.right.right.left = newNode(7);`   `// Print all nodes having` `// children as their factors` `getCount(root);` `console.log(count);`

Output

`3`

Time Complexity: O(N) where N is the number of nodes in given binary tree.

Auxiliary Space: O(N) due to recursion call stack.

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