Skip to content
Related Articles

Related Articles

Improve Article

Count of nodes in a Binary Tree whose child is its prime factors

  • Difficulty Level : Medium
  • Last Updated : 18 Jun, 2021

Given a Binary Tree, the task is to print the count of nodes whose either of the immediate child is its prime factors
Examples: 
 

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

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

 

Approach: 
 

  1. Traverse the given Binary Tree and for each node: 
    • Check if children exist or not.
    • If the children exist, check if each child is a prime factor of this node or not. 
    • Keep the count of such nodes and print it at the end.
       
  2. In order to check if a factor is prime, we will use Sieve of Eratosthenes to precompute the prime numbers to do the checking in O(1). 
     

Below is the implementation of the above approach:
 

C++




// C++ program for Counting nodes whose
// immediate children are its factors
 
#include <bits/stdc++.h>
using namespace std;
 
int N = 1000000;
 
// To store all prime numbers
vector<int> prime;
 
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..N]"
    // and initialize all its entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    bool check[N + 1];
    memset(check, true, sizeof(check));
 
    for (int p = 2; p * p <= N; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
 
            prime.push_back(p);
 
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
 
// 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 if immediate children
// of a node are its factors or not
bool IsChilrenPrimeFactor(struct Node* parent,
                          struct Node* a)
{
    if (prime[a->key]
        && (parent->key % a->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
 
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.push(node);
 
    while (!q.empty()) {
        struct Node* temp = q.front();
        q.pop();
 
        // If only right child exist
        if (temp->left == NULL
            && temp->right != NULL) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp->right))
                count++;
        }
        else {
            // If only left child exist
            if (temp->right == NULL
                && temp->left != NULL) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp->left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp->left != NULL
                    && temp->right != NULL) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp->right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp->left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp->left != NULL)
            q.push(temp->left);
 
        // Check for right child
        if (temp->right != NULL)
            q.push(temp->right);
    }
    return count;
}
 
// Driver Code
int main()
{
    /*       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   14
                      /
                     7
    */
 
    // Create Binary Tree as shown
    Node* root = newNode(10);
 
    root->left = newNode(2);
    root->right = newNode(5);
 
    root->right->left = newNode(18);
    root->right->right = newNode(12);
 
    root->right->left->left = newNode(2);
    root->right->left->right = newNode(3);
    root->right->right->left = newNode(3);
    root->right->right->right = newNode(14);
    root->right->right->right->left = newNode(7);
 
    // To save all prime numbers
    SieveOfEratosthenes();
 
    // Print Count of 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{
  
static int N = 1000000;
  
// To store all prime numbers
static Vector<Integer> prime = new Vector<Integer>();
  
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..N]"
    // and initialize all its entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    boolean []check = new boolean[N + 1];
    Arrays.fill(check, true);
  
    for (int p = 2; p * p <= N; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
  
            prime.add(p);
  
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
  
// 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 if immediate children
// of a node are its factors or not
static boolean IsChilrenPrimeFactor(Node parent,
                          Node a)
{
    if (prime.get(a.key)>0
        && (parent.key % a.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<>();
  
    // Do level order traversal starting
    // from root
  
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.add(node);
  
    while (!q.isEmpty()) {
        Node temp = q.peek();
        q.remove();
  
        // If only right child exist
        if (temp.left == null
            && temp.right != null) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp.right))
                count++;
        }
        else {
            // If only left child exist
            if (temp.right == null
                && temp.left != null) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp.left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp.left != null
                    && temp.right != null) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp.right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp.left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp.left != null)
            q.add(temp.left);
  
        // Check for right child
        if (temp.right != null)
            q.add(temp.right);
    }
    return count;
}
  
// Driver Code
public static void main(String[] args)
{
    /*       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   14
                      /
                     7
    */
  
    // Create Binary Tree as shown
    Node root = newNode(10);
  
    root.left = newNode(2);
    root.right = newNode(5);
  
    root.right.left = newNode(18);
    root.right.right = newNode(12);
  
    root.right.left.left = newNode(2);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(3);
    root.right.right.right = newNode(14);
    root.right.right.right.left = newNode(7);
  
    // To save all prime numbers
    SieveOfEratosthenes();
  
    // Print Count of all nodes having children
    // as their factors
    System.out.print(GetCount(root) +"\n");
  
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program for Counting nodes whose
# immediate children are its factors
from collections import deque
N = 1000000
 
# To store all prime numbers
prime = []
def SieveOfEratosthenes() -> None:
 
    # Create a boolean array "prime[0..N]"
    # and initialize all its entries as true.
    # A value in prime[i] will finally
    # be false if i is Not a prime, else true.
    check = [True for _ in range(N + 1)]
    p = 2
    while p * p <= N:
       
        # If prime[p] is not changed,
        # then it is a prime
        if (check[p]):
            prime.append(p)
 
            # Update all multiples of p
            # greater than or equal to
            # the square of it
            # numbers which are multiples of p
            # and are less than p^2
            # are already marked.
            for i in range(p * p, N + 1, p):
                check[i] = False
        p += 1
 
# A Tree node
class Node:
    def __init__(self, key: int) -> None:
        self.key = key
        self.left = None
        self.right = None
 
# Function to check if immediate children
# of a node are its factors or not
def IsChilrenPrimeFactor(parent: Node, a: Node) -> bool:
    if (prime[a.key] and (parent.key % a.key == 0)):
        return True
    else:
        return False
 
# Function to get the count of full Nodes in
# a binary tree
def GetCount(node: Node) -> int:
 
    # If tree is empty
    if (not node):
        return 0
    q = deque()
 
    # Do level order traversal starting
    # from root
 
    # Initialize count of full nodes having
    # children as their factors
    count = 0
    q.append(node)
 
    while q:
        temp = q.popleft()
 
        # If only right child exist
        if (temp.left == None and temp.right != None):
            if (IsChilrenPrimeFactor(temp, temp.right)):
                count += 1
 
        else:
           
            # If only left child exist
            if (temp.right == None and temp.left != None):
                if (IsChilrenPrimeFactor(temp, temp.left)):
                    count += 1
 
            else:
               
                # Both left and right child exist
                if (temp.left != None and temp.right != None):
                    if (IsChilrenPrimeFactor(temp, temp.right)
                            and IsChilrenPrimeFactor(temp, temp.left)):
                        count += 1
 
        # Check for left child
        if (temp.left != None):
            q.append(temp.left)
 
        # Check for right child
        if (temp.right != None):
            q.append(temp.right)
 
    return count
 
# Driver Code
if __name__ == "__main__":
    '''       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   14
                      /
                     7
    '''
 
    # Create Binary Tree as shown
    root = Node(10)
 
    root.left = Node(2)
    root.right = Node(5)
 
    root.right.left = Node(18)
    root.right.right = Node(12)
 
    root.right.left.left = Node(2)
    root.right.left.right = Node(3)
    root.right.right.left = Node(3)
    root.right.right.right = Node(14)
    root.right.right.right.left = Node(7)
 
    # To save all prime numbers
    SieveOfEratosthenes()
 
    # Print Count of all nodes having children
    # as their factors
    print(GetCount(root))
 
# This code is contributed by sanjeev2552

C#




// C# program for Counting nodes whose
// immediate children are its factors
using System;
using System.Collections.Generic;
 
public class GFG{
   
static int N = 1000000;
   
// To store all prime numbers
static List<int> prime = new List<int>();
   
static void SieveOfEratosthenes()
{
    // Create a bool array "prime[0..N]"
    // and initialize all its entries as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    bool []check = new bool[N + 1];
    for (int i = 0; i <= N; i += 1)
        check[i] = true;
   
    for (int p = 2; p * p <= N; p++) {
   
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
   
            prime.Add(p);
   
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
   
// 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 if immediate children
// of a node are its factors or not
static bool IsChilrenPrimeFactor(Node parent,
                          Node a)
{
    if (prime[a.key]>0
        && (parent.key % a.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
   
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.Add(node);
   
    while (q.Count!=0) {
        Node temp = q[0];
        q.RemoveAt(0);
   
        // If only right child exist
        if (temp.left == null
            && temp.right != null) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp.right))
                count++;
        }
        else {
            // If only left child exist
            if (temp.right == null
                && temp.left != null) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp.left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp.left != null
                    && temp.right != null) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp.right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp.left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp.left != null)
            q.Add(temp.left);
   
        // Check for right child
        if (temp.right != null)
            q.Add(temp.right);
    }
    return count;
}
   
// Driver Code
public static void Main(String[] args)
{
    /*       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   14
                      /
                     7
    */
   
    // Create Binary Tree as shown
    Node root = newNode(10);
   
    root.left = newNode(2);
    root.right = newNode(5);
   
    root.right.left = newNode(18);
    root.right.right = newNode(12);
   
    root.right.left.left = newNode(2);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(3);
    root.right.right.right = newNode(14);
    root.right.right.right.left = newNode(7);
   
    // To save all prime numbers
    SieveOfEratosthenes();
   
    // Print Count of all nodes having children
    // as their factors
    Console.Write(GetCount(root) +"\n");
   
}
}
// This code contributed by Rajput-Ji

Javascript




<script>
 
  // JavaScript program for Counting nodes whose
  // immediate children are its factors
   
  let N = 1000000;
   
  // To store all prime numbers
  let prime = [];
     
  function SieveOfEratosthenes()
  {
      // Create a boolean array "prime[0..N]"
      // and initialize all its entries as true.
      // A value in prime[i] will finally
      // be false if i is Not a prime, else true.
      let check = new Array(N + 1);
      check.fill(true);
     
      for (let p = 2; p * p <= N; p++) {
     
          // If prime[p] is not changed,
          // then it is a prime
          if (check[p] == true) {
     
              prime.push(p);
     
              // Update all multiples of p
              // greater than or equal to
              // the square of it
              // numbers which are multiples of p
              // and are less than p^2
              // are already marked.
              for (let i = p * p; i <= N; i += p)
                  check[i] = false;
          }
      }
  }
   
  // A Tree node
  class Node
  {
      constructor(key) {
         this.left = null;
         this.right = null;
         this.key = key;
      }
  }
     
  // Utility function to create a new node
  function newNode(key)
  {
      let temp = new Node(key);
      return (temp);
  }
     
  // Function to check if immediate children
  // of a node are its factors or not
  function IsChilrenPrimeFactor(parent, a)
  {
      if (prime[a.key]>0
          && (parent.key % a.key == 0))
          return true;
      else
          return false;
  }
     
  // Function to get the count of full Nodes in
  // a binary tree
  function GetCount(node)
  {
      // If tree is empty
      if (node == null)
          return 0;
      let q = [];
     
      // Do level order traversal starting
      // from root
     
      // Initialize count of full nodes having
      // children as their factors
      let count = 0;
      q.push(node);
     
      while (q.length > 0) {
          let temp = q[0];
          q.shift();
     
          // If only right child exist
          if (temp.left == null
              && temp.right != null) {
              if (IsChilrenPrimeFactor(
                      temp,
                      temp.right))
                  count++;
          }
          else {
              // If only left child exist
              if (temp.right == null
                  && temp.left != null) {
                  if (IsChilrenPrimeFactor(
                          temp,
                          temp.left))
                      count++;
              }
              else {
                  // Both left and right child exist
                  if (temp.left != null
                      && temp.right != null) {
                      if (IsChilrenPrimeFactor(
                              temp,
                              temp.right)
                          && IsChilrenPrimeFactor(
                                 temp,
                                 temp.left))
                          count++;
                  }
              }
          }
          // Check for left child
          if (temp.left != null)
              q.push(temp.left);
     
          // Check for right child
          if (temp.right != null)
              q.push(temp.right);
      }
      return count;
  }
   
  /*       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   14
                      /
                     7
    */
   
  // Create Binary Tree as shown
  let root = newNode(10);
 
  root.left = newNode(2);
  root.right = newNode(5);
 
  root.right.left = newNode(18);
  root.right.right = newNode(12);
 
  root.right.left.left = newNode(2);
  root.right.left.right = newNode(3);
  root.right.right.left = newNode(3);
  root.right.right.right = newNode(14);
  root.right.right.right.left = newNode(7);
 
  // To save all prime numbers
  SieveOfEratosthenes();
 
  // Print Count of all nodes having children
  // as their factors
  document.write(GetCount(root) +"</br>");
     
</script>
Output: 



3

 

Complexity Analysis: 
 

  • Time Complexity : O(N). 
    In dfs, every node of the tree is processed once and hence the complexity due to the bfs is O(N) if there are total N nodes in the tree. Also, for processing each node the SieveOfEratosthenes() function is used which has a complexity of O(sqrt(N)) too but since this function is executed only once, it does not affect the overall time complexity. Therefore, the time complexity is O(N).
  • Auxiliary Space : O(N). 
    Extra space is used for the prime array, so the space complexity is O(N).

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :