Count of exponential paths in a Binary Tree
Last Updated :
12 Jul, 2025
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:
- First find the value of x for which xy=root & x is minimum possible & y>0.
- If current value of the node is not equal to xy 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++
// 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 recursive 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;
}
// 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 recursive 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
# Python3 program to find the count
# exponential paths in Binary Tree
import math
# Structure of a Tree node
class Node:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# Function to create a new node
def newNode(key):
temp = Node(key)
return temp
# Function to find x
def find_x(n):
if n == 1:
return 1
# Take log10 of n
num = math.log10(n)
x, no = 0, 0
for i in range(2, n + 1):
den = math.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
def is_key(n, x):
# Take logx(n) with base x
p = math.log10(n) / math.log10(x)
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
def evenPaths(node, count, x):
# Base Condition, when node pointer
# becomes null or node value is not
# a number of pow(x, y )
if node == None or not is_key(node.key, x):
return count
# Increment count when
# encounter leaf node
if node.left == None and node.right == None:
count+=1
# Left recursive call
# save the value of count
count = evenPaths(node.left, count, x)
# Right recursive call and
# return value of count
return evenPaths(node.right, count, x)
# Function to count exponential paths
def countExpPaths(node, x):
return evenPaths(node, 0, x)
# Create Tree
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
x = find_x(root.key)
# Function call
print(countExpPaths(root, x))
# This code is contributed by divyeshrabadiya07.
// 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 recursive 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
<script>
// Javascript program to find the count
// exponential paths in Binary Tree
// Structure of a Tree node
class Node
{
constructor(key)
{
this.left = null;
this.right = null;
this.key = key;
}
}
// Function to create a new node
function newNode(key)
{
let temp = new Node(key);
return (temp);
}
// Function to find x
function find_x(n)
{
if (n == 1)
return 1;
let num, den, p;
// Take log10 of n
num = Math.log10(n);
let x = 0, no = 0;
for(let i = 2; i <= n; i++)
{
den = Math.log10(i);
// Log(n) with base i
p = num / den;
// Raising i to the power p
no = parseInt(Math.pow(
i, parseInt(p, 10)), 10);
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
function is_key(n, x)
{
let p;
// Take logx(n) with base x
p = Math.log10(n) / Math.log10(x);
let no = parseInt(Math.pow(
x, parseInt(p, 10)), 10);
if (n == no)
return true;
return false;
}
// Utility function to count
// the exponent path in a
// given Binary tree
function evenPaths(node, count, 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 recursive call and
// return value of count
return evenPaths(node.right,
count, x);
}
// Function to count exponential paths
function countExpPaths(node, x)
{
return evenPaths(node, 0, x);
}
// Driver code
// Create Tree
let 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
let x = find_x(root.key);
// Function call
document.write(countExpPaths(root, x));
// This code is contributed by mukesh07
</script>
Time Complexity: O(n log n) as it contains a loop that runs from 2 to n, and for each i, it calculates log10(i) and performs a division and a power operation. The time complexity of the function is_key is O(log n) as it calculates log10(n) / log10(x) and performs a power operation. The time complexity of the function evenPaths is O(n) as it traverses each node in the binary tree once
Auxiliary Space: O(h), where h is the height of the binary tree. This is because the recursive functions evenPaths and countExpPaths use the call stack to store the function call frames, and the maximum number of function calls on the call stack is equal to the height of the binary tree. Additionally, the program uses O(1) extra space to store variables and O(n) space to store the binary tree.
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