Skip to content
Related Articles

Related Articles

Count of exponential paths in a Binary Tree
  • Last Updated : 01 Dec, 2020

Given a Binary Tree, the task is to count the number of Exponential paths in the given Binary Tree.  

Exponential Path is a path where root to leaf path contains all nodes being equal to xy, & where x is a minimum possible positive constant & y is a variable positive integer.

Example:  

Input:
             27
           /    \
          9      81
         / \    /  \
        3  10  70   243
                   /   \
                  81   909
Output: 2 
Explanation:
There are 2 exponential path for
the above Binary Tree, for x = 3,
Path 1: 27 -> 9 -> 3
Path 2: 27 -> 81 -> 243 -> 81

Input:
             8
           /    \
          4      81
         / \    /  \
        3   2  70   243
                   /   \
                  81   909
Output: 1

Approach: The idea is to use Preorder Tree Traversal. During preorder traversal of the given binary tree do the following:  

  1. First find the value of x for which xy=root & x is minimum possible & y>0.
  2. If current value of the node is not equal to xy for some y>0, or pointer becomes NULL then return the count.
  3. If the current node is a leaf node then increment the count by 1.
  4. Recursively call for the left and right subtree with the updated count.
  5. After all recursive call, the value of count is number of exponential paths for a given binary tree.

Below is the implementation of the above approach: 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the count
// exponential paths in Binary Tree
 
#include <bits/stdc++.h>
using namespace std;
 
// A Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
 
// 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 find x
int find_x(int n)
{
    if (n == 1)
        return 1;
 
    double num, den, p;
 
    // Take log10 of n
    num = log10(n);
 
    int x, no;
 
    for (int i = 2; i <= n; i++) {
        den = log10(i);
 
        // Log(n) with base i
        p = num / den;
 
        // Raising i to the power p
        no = (int)(pow(i, int(p)));
 
        if (abs(no - n) < 1e-6) {
            x = i;
            break;
        }
    }
 
    return x;
}
 
// function To check
// whether the given node
// equals to x^y for some y>0
bool is_key(int n, int x)
{
    double p;
 
    // Take logx(n) with base x
    p = log10(n) / log10(x);
 
    int no = (int)(pow(x, int(p)));
 
    if (n == no)
        return true;
 
    return false;
}
 
// Utility function to count
// the exponent path
// in a given Binary tree
int evenPaths(struct Node* node,
              int count, int x)
{
 
    // Base Condition, when node pointer
    // becomes null or node value is not
    // a number of pow(x, y )
    if (node == NULL
        || !is_key(node->key, x)) {
        return count;
    }
 
    // Increment count when
    // encounter leaf node
    if (!node->left
        && !node->right) {
        count++;
    }
 
    // Left recursive call
    // save the value of count
    count = evenPaths(
        node->left, count, x);
 
    // Right reursive call and
    // return value of count
    return evenPaths(
        node->right, count, x);
}
 
// function to count exponential paths
int countExpPaths(
    struct Node* node, int x)
{
    return evenPaths(node, 0, x);
}
 
// Driver Code
int main()
{
 
    // create Tree
    Node* root = newNode(27);
    root->left = newNode(9);
    root->right = newNode(81);
 
    root->left->left = newNode(3);
    root->left->right = newNode(10);
 
    root->right->left = newNode(70);
    root->right->right = newNode(243);
    root->right->right->left = newNode(81);
    root->right->right->right = newNode(909);
 
    // retrieve the value of x
    int x = find_x(root->key);
 
    // Function call
    cout << countExpPaths(root, x);
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the count
// exponential paths in Binary Tree
import java.util.*;
import java.lang.*;
 
class GFG{
   
// Structure of a Tree node
static class Node
{
    int key;
    Node left, right;
}
  
// 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 find x
static int find_x(int n)
{
    if (n == 1)
        return 1;
   
    double num, den, p;
   
    // Take log10 of n
    num = Math.log10(n);
   
    int x = 0, no = 0;
     
    for(int i = 2; i <= n; i++)
    {
        den = Math.log10(i);
         
        // Log(n) with base i
        p = num / den;
         
        // Raising i to the power p
        no = (int)(Math.pow(i, (int)p));
   
        if (Math.abs(no - n) < 1e-6)
        {
            x = i;
            break;
        }
    }
    return x;
}
   
// Function to check whether the
// given node equals to x^y for some y>0
static boolean is_key(int n, int x)
{
    double p;
     
    // Take logx(n) with base x
    p = Math.log10(n) / Math.log10(x);
   
    int no = (int)(Math.pow(x, (int)p));
   
    if (n == no)
        return true;
   
    return false;
}
   
// Utility function to count
// the exponent path in a
// given Binary tree
static int evenPaths(Node node, int count,
                                int x)
{
     
    // Base Condition, when node pointer
    // becomes null or node value is not
    // a number of pow(x, y )
    if (node == null || !is_key(node.key, x))
    {
        return count;
    }
   
    // Increment count when
    // encounter leaf node
    if (node.left == null &&
       node.right == null)
    {
        count++;
    }
     
    // Left recursive call
    // save the value of count
    count = evenPaths(node.left,
                      count, x);
   
    // Right reursive call and
    // return value of count
    return evenPaths(node.right,
                     count, x);
}
   
// Function to count exponential paths
static int countExpPaths(Node node, int x)
{
    return evenPaths(node, 0, x);
 
// Driver code
public static void main(String[] args)
{
     
    // Create Tree
    Node root = newNode(27);
    root.left = newNode(9);
    root.right = newNode(81);
     
    root.left.left = newNode(3);
    root.left.right = newNode(10);
     
    root.right.left = newNode(70);
    root.right.right = newNode(243);
    root.right.right.left = newNode(81);
    root.right.right.right = newNode(909);
     
    // Retrieve the value of x
    int x = find_x(root.key);
     
    // Function call
    System.out.println(countExpPaths(root, x));
}
}
 
// This code is contributed by offbeat

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the count
// exponential paths in Binary Tree
using System;
 
class GFG{
    
// Structure of a Tree node
public class Node
{
    public int key;
    public Node left, right;
}
   
// 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 find x
static int find_x(int n)
{
    if (n == 1)
        return 1;
         
    double num, den, p;
    
    // Take log10 of n
    num = Math.Log10(n);
    
    int x = 0, no = 0;
      
    for(int i = 2; i <= n; i++)
    {
        den = Math.Log10(i);
          
        // Log(n) with base i
        p = num / den;
          
        // Raising i to the power p
        no = (int)(Math.Pow(i, (int)p));
    
        if (Math.Abs(no - n) < 0.000001)
        {
            x = i;
            break;
        }
    }
    return x;
}
    
// Function to check whether the
// given node equals to x^y for some y>0
static bool is_key(int n, int x)
{
    double p;
      
    // Take logx(n) with base x
    p = Math.Log10(n) / Math.Log10(x);
    
    int no = (int)(Math.Pow(x, (int)p));
    
    if (n == no)
        return true;
    
    return false;
}
    
// Utility function to count
// the exponent path in a
// given Binary tree
static int evenPaths(Node node, int count,
                                int x)
{
     
    // Base Condition, when node pointer
    // becomes null or node value is not
    // a number of pow(x, y )
    if (node == null || !is_key(node.key, x))
    {
        return count;
    }
    
    // Increment count when
    // encounter leaf node
    if (node.left == null &&
       node.right == null)
    {
        count++;
    }
      
    // Left recursive call
    // save the value of count
    count = evenPaths(node.left,
                      count, x);
    
    // Right reursive call and
    // return value of count
    return evenPaths(node.right,
                     count, x);
}
    
// Function to count exponential paths
static int countExpPaths(Node node, int x)
{
    return evenPaths(node, 0, x);
  
// Driver code
public static void Main(string[] args)
{
     
    // Create Tree
    Node root = newNode(27);
    root.left = newNode(9);
    root.right = newNode(81);
      
    root.left.left = newNode(3);
    root.left.right = newNode(10);
      
    root.right.left = newNode(70);
    root.right.right = newNode(243);
    root.right.right.left = newNode(81);
    root.right.right.right = newNode(909);
      
    // Retrieve the value of x
    int x = find_x(root.key);
      
    // Function call
    Console.Write(countExpPaths(root, x));
}
}
 
// This code is contributed by rutvik_56

chevron_right


Output: 

2

 

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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :