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Count of nodes in a Binary Tree whose immediate children are co-prime

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Given a Binary Tree, the task is to count the nodes whose immediate children are co-prime. 

Examples: 

Input: 
                  1
                /   \ 
              15     5
             /  \   /  \ 
            11   2 4     15 
                    \    / 
                     2  3  
Output: 2
Explanation: 
Children of 15 (11, 2) are co-prime
Children of 5 (4, 15) are co-prime

Input:
                 7
                /  \ 
              21     14 
             /  \      \
            77   16     3 
           / \         / \
          2   5       10  11
                      /
                     23 
Output:3
Explanation: 
Children of 21 (77, 8) are co-prime
Children of 77 (2, 5) are co-prime
Children of 3 (10, 11) are co-prime

Approach: The idea is to: 

  1. Do level order traversal of the tree
  2. For each node check that its both children are not Null
  3. If true, then check whether greatest common divisor of both children is 1.
  4. If yes, then count such nodes and print at the end.

Below is the implementation of the above approach:

C++




// C++ program for Counting nodes
// whose immediate children
// are co-prime
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of 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
// two nodes are co-prime or not
bool coprime(struct Node* a,
             struct Node* b)
{
 
    if (__gcd(a->key, b->key) == 1)
        return true;
    else
        return false;
}
 
// Function to get the count of
// Nodes whose immediate children
// are co-prime in a binary tree
unsigned int getCount(struct Node* node)
{
    // Base Case
    if (!node)
        return 0;
 
    queue<Node*> q;
 
    // Do level order traversal
    // starting from root
    int count = 0;
    q.push(node);
 
    while (!q.empty()) {
        struct Node* temp = q.front();
        q.pop();
 
        if (temp->left && temp->right) {
            if (coprime(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
// number 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);
}
 
// Function to create Tree
// and find the count of nodes
// whose immediate children
// are co-prime
void findCount()
{
    /*         10
            /  \
          48   12
              /  \
            18    35
           / \    / \
          21 29  43 16
                 /
                7
    */
 
    // Create Binary Tree
    Node* root = newNode(10);
    root->left = newNode(48);
    root->right = newNode(12);
 
    root->right->left = newNode(18);
    root->right->right = newNode(35);
 
    root->right->left->left = newNode(21);
    root->right->left->right = newNode(29);
    root->right->right->left = newNode(43);
    root->right->right->right = newNode(16);
    root->right->right->right->left = newNode(7);
 
    // Print all nodes
    // with Co-Prime children
    cout << getCount(root) << endl;
}
 
// Driver Code
int main()
{
    // Function Call
    findCount();
 
    return 0;
}

Java



// Java program for Counting nodes
// whose immediate children
// are co-prime
import java.util.*;
 
class GFG{
  
// Structure of 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
// two nodes are co-prime or not
static boolean coprime(Node a,
             Node b)
{
  
    if (__gcd(a.key, b.key) == 1)
        return true;
    else
        return false;
}
  
// Function to get the count of
// Nodes whose immediate children
// are co-prime in a binary tree
static int getCount(Node node)
{
    // Base Case
    if (node == null)
        return 0;
  
    Queue<Node> q = new LinkedList<Node>();
  
    // Do level order traversal
    // starting from root
    int count = 0;
    q.add(node);
  
    while (!q.isEmpty()) {
        Node temp = q.peek();
        q.remove();
  
        if (temp.left != null && temp.right != null) {
            if (coprime(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
// number 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);
}
  
// Function to create Tree
// and find the count of nodes
// whose immediate children
// are co-prime
static void findCount()
{
    /*         10
            /  \
          48   12
              /  \
            18    35
           / \    / \
          21 29  43 16
                 /
                7
    */
  
    // Create Binary Tree
    Node root = newNode(10);
    root.left = newNode(48);
    root.right = newNode(12);
  
    root.right.left = newNode(18);
    root.right.right = newNode(35);
  
    root.right.left.left = newNode(21);
    root.right.left.right = newNode(29);
    root.right.right.left = newNode(43);
    root.right.right.right = newNode(16);
    root.right.right.right.left = newNode(7);
  
    // Print all nodes
    // with Co-Prime children
    System.out.print(getCount(root) +"\n");
}
static int __gcd(int a, int b) 
{ 
    return b == 0? a:__gcd(b, a % b);    
}
 
// Driver Code
public static void main(String[] args)
{
    // Function Call
    findCount();
  
}
}
 
// This code is contributed by sapnasingh4991
Python3



# Python3 program for counting nodes
# whose immediate children
# are co-prime
from collections import deque as queue
from math import gcd as __gcd
 
# 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
# two nodes are co-prime or not
def coprime(a, b):
     
    if (__gcd(a.data, b.data) == 1):
        return True
    else:
        return False
 
# Function to get the count of
# Nodes whose immediate children
# are co-prime 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
    q.append(node)
 
    while (len(q) > 0):
        temp = q.popleft()
        #q.pop()
 
        if (temp.left and temp.right):
            if (coprime(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))
 
# Function to create Tree
# and find the count of nodes
# whose immediate children
# are co-prime
def findCount():
     
    #          10
    #         /  \
    #       48   12
    #           /  \
    #         18    35
    #        / \    / \
    #       21 29  43 16
    #              /
    #             7
    #
 
    # Create Binary Tree
    root = Node(10)
    root.left = Node(48)
    root.right = Node(12)
 
    root.right.left = Node(18)
    root.right.right = Node(35)
 
    root.right.left.left = Node(21)
    root.right.left.right = Node(29)
    root.right.right.left = Node(43)
    root.right.right.right = Node(16)
    root.right.right.right.left = Node(7)
 
    # Print all nodes
    # with Co-Prime children
    print(getCount(root))
 
# Driver Code
if __name__ == '__main__':
     
    # Function Call
    findCount()
 
# This code is contributed by mohit kumar 29
C#



// C# program for Counting nodes
// whose immediate children
// are co-prime
using System;
using System.Collections.Generic;
 
class GFG{
   
// Structure of 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
// two nodes are co-prime or not
static bool coprime(Node a,
             Node b)
{
   
    if (__gcd(a.key, b.key) == 1)
        return true;
    else
        return false;
}
   
// Function to get the count of
// Nodes whose immediate children
// are co-prime in a binary tree
static int getCount(Node node)
{
    // Base Case
    if (node == null)
        return 0;
   
    List<Node> q = new List<Node>();
   
    // Do level order traversal
    // starting from root
    int count = 0;
    q.Add(node);
   
    while (q.Count != 0) {
        Node temp = q[0];
        q.RemoveAt(0);
   
        if (temp.left != null && temp.right != null) {
            if (coprime(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
// number 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);
}
   
// Function to create Tree
// and find the count of nodes
// whose immediate children
// are co-prime
static void findCount()
{
    /*         10
            /  \
          48   12
              /  \
            18    35
           / \    / \
          21 29  43 16
                 /
                7
    */
   
    // Create Binary Tree
    Node root = newNode(10);
    root.left = newNode(48);
    root.right = newNode(12);
   
    root.right.left = newNode(18);
    root.right.right = newNode(35);
   
    root.right.left.left = newNode(21);
    root.right.left.right = newNode(29);
    root.right.right.left = newNode(43);
    root.right.right.right = newNode(16);
    root.right.right.right.left = newNode(7);
   
    // Print all nodes
    // with Co-Prime children
    Console.Write(getCount(root) +"\n");
}
static int __gcd(int a, int b) 
{ 
    return b == 0? a:__gcd(b, a % b);    
}
  
// Driver Code
public static void Main(String[] args)
{
    // Function Call
    findCount();
}
}
 
// This code is contributed by 29AjayKumar
Javascript



<script>
 
// Javascript program for counting nodes
// whose immediate children are co-prime
 
// Structure of node
class Node
{
     
    // Utility function to
    // create a new node
    constructor(key)
    {
        this.key = key;
        this.left = this.right = null;
    }
}
 
// Function to check and print if
// two nodes are co-prime or not
function coprime(a, b)
{
    if (__gcd(a.key, b.key) == 1)
        return true;
    else
        return false;
}
 
// Function to get the count of
// Nodes whose immediate children
// are co-prime in a binary tree
function getCount(node)
{
     
    // Base Case
    if (node == null)
        return 0;
   
    let q = [];
   
    // Do level order traversal
    // starting from root
    let count = 0;
    q.push(node);
   
    while (q.length != 0)
    {
        let temp = q.shift();
         
        if (temp.left != null &&
            temp.right != null)
        {
            if (coprime(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
// number of nodes
// In a given binary tree
function findSize(node)
{
     
    // Base condition
    if (node == null)
        return 0;
   
    return 1 + findSize(node.left) +
               findSize(node.right);
}
 
// Function to create Tree
// and find the count of nodes
// whose immediate children
// are co-prime
function findCount()
{
     
    /*       10
            /  \
          48   12
              /  \
            18    35
           / \    / \
          21 29  43 16
                 /
                7
    */
   
    // Create Binary Tree
    let root = new Node(10);
    root.left = new Node(48);
    root.right = new Node(12);
   
    root.right.left = new Node(18);
    root.right.right = new Node(35);
   
    root.right.left.left = new Node(21);
    root.right.left.right = new Node(29);
    root.right.right.left = new Node(43);
    root.right.right.right = new Node(16);
    root.right.right.right.left = new Node(7);
   
    // Print all nodes
    // with Co-Prime children
    document.write(getCount(root) + "<br>");
}
 
function  __gcd(a,b)
{
    return b == 0? a:__gcd(b, a % b);   
}
 
// Driver Code
 
// Function Call
findCount();
 
// This code is contributed by patel2127
 
</script>
Output: 
3
 
Complexity Analysis:
Time Complexity: O(N*logV) where V is the weight of a node in the tree.
In bfs, 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, while processing every node, in order to check if the node values are co-prime or not, the inbuilt __gcd(A, B) function where A, B are the weight of the nodes is being called and this function has a complexity of O(log(min(A, B))), hence for every node there is an added complexity of O(logV). Therefore, the time complexity is O(N*logV).
Auxiliary Space : O(w) where w is the maximum width of the tree.

Last Updated :
04 Feb, 2022
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