Count of exponential paths in a Binary Tree

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 x^y, & 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:

First find the value of x for which x^y=root & x is minimum possible & y>0 .
If current value of the node is not equal to x^y for some y>0, or pointer becomes NULL then return the count.
If the current node is a leaf node then increment the count by 1.
Recursively call for the left and right subtree with the updated count.
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++ 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; 
} 

Output:

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