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
// 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
<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> |
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.
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