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


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++

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// C++ program for Counting nodes
// whose immediate children
// are its factors
 
#include <bits/stdc++.h>
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<Node*> 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;
}

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Java

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// 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<Node> q = new LinkedList<Node>();
  
    // 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

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Python3

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# 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

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C#

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// 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<Node> q = new List<Node>();
   
    // 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

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Output: 

3




 

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