# Count of exponential paths in a Binary Tree

Last Updated : 27 Mar, 2023

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++

 `// C++ program to find the count` `// exponential paths in Binary Tree`   `#include ` `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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`2`

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.