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

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
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

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 `

Output:

```3
```

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