Number of Binary Search Trees of height H consisting of H+1 nodes
Given a positive integer H, the task is to find the number of possible Binary Search Trees of height H consisting of the first (H + 1) natural numbers as the node values. Since the count can be very large, print it to modulo 109 + 7.
Examples:
Input: H = 2
Output: 4
Explanation: All possible BSTs of height 2 consisting of 3 nodes are as follows:
Therefore, the total number of BSTs possible is 4.
Input: H = 6
Output: 64
Approach: The given problem can be solved based on the following observations:
- Only (H + 1) nodes are can be used to form a Binary Tree of height H.
- Except for the root node, every node has two possibilities, i.e. either to be the left child or to be the right child.
- Considering T(H) to be the number of BST of height H, where T(0) = 1 and T(H) = 2 * T(H – 1).
- Solving the above recurrence relation, the value of T(H) is 2H.
Therefore, from the above observations, print the value of 2H as the total number of BSTs of height H consisting of the first (H + 1) natural numbers.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int mod = 1000000007;// Function to calculate x^y// modulo 1000000007 in O(log y)int power(long long x, unsigned int y){ // Stores the value of x^y int res = 1; // Update x if it exceeds mod x = x % mod; // If x is divisible by mod if (x == 0) return 0; while (y > 0) { // If y is odd, then // multiply x with result if (y & 1) res = (res * x) % mod; // Divide y by 2 y = y >> 1; // Update the value of x x = (x * x) % mod; } // Return the value of x^y return res;}// Function to count the number of// of BSTs of height H consisting// of (H + 1) nodesint CountBST(int H){ return power(2, H);}// Driver Codeint main(){ int H = 2; cout << CountBST(H); return 0;} |
Java
// Java program for the above approachclass GFG{ static int mod = 1000000007;// Function to calculate x^y// modulo 1000000007 in O(log y)static long power(long x, int y){ // Stores the value of x^y long res = 1; // Update x if it exceeds mod x = x % mod; // If x is divisible by mod if (x == 0) return 0; while (y > 0) { // If y is odd, then // multiply x with result if ((y & 1) == 1) res = (res * x) % mod; // Divide y by 2 y = y >> 1; // Update the value of x x = (x * x) % mod; } // Return the value of x^y return res;}// Function to count the number of// of BSTs of height H consisting// of (H + 1) nodesstatic long CountBST(int H) { return power(2, H); }// Driver codepublic static void main(String[] args){ int H = 2; System.out.print(CountBST(H));}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to calculate x^y# modulo 1000000007 in O(log y)def power(x, y): mod = 1000000007 # Stores the value of x^y res = 1 # Update x if it exceeds mod x = x % mod # If x is divisible by mod if (x == 0): return 0 while (y > 0): # If y is odd, then # multiply x with result if (y & 1): res = (res * x) % mod # Divide y by 2 y = y >> 1 # Update the value of x x = (x * x) % mod # Return the value of x^y return res# Function to count the number of# of BSTs of height H consisting# of (H + 1) nodesdef CountBST(H): return power(2, H)# Driver CodeH = 2print(CountBST(H))# This code is contributed by rohitsingh07052 |
C#
// C# program for the above approach using System;class GFG{ static int mod = 1000000007;// Function to calculate x^y// modulo 1000000007 in O(log y)static long power(long x, int y){ // Stores the value of x^y long res = 1; // Update x if it exceeds mod x = x % mod; // If x is divisible by mod if (x == 0) return 0; while (y > 0) { // If y is odd, then // multiply x with result if ((y & 1) == 1) res = (res * x) % mod; // Divide y by 2 y = y >> 1; // Update the value of x x = (x * x) % mod; } // Return the value of x^y return res;}// Function to count the number of// of BSTs of height H consisting// of (H + 1) nodesstatic long CountBST(int H){ return power(2, H);}// Driver codestatic void Main(){ int H = 2; Console.Write(CountBST(H));}}// This code is contributed by abhinavjain194 |
Javascript
<script>// Javascript program for the above approachvar mod = 1000000007;// Function to calculate x^y// modulo 1000000007 in O(log y)function power(x, y){ // Stores the value of x^y var res = 1; // Update x if it exceeds mod x = x % mod; // If x is divisible by mod if (x == 0) return 0; while (y > 0) { // If y is odd, then // multiply x with result if (y & 1) res = (res * x) % mod; // Divide y by 2 y = y >> 1; // Update the value of x x = (x * x) % mod; } // Return the value of x^y return res;}// Function to count the number of// of BSTs of height H consisting// of (H + 1) nodesfunction CountBST(H){ return power(2, H);}// Driver Codevar H = 2;document.write( CountBST(H));</script> |
Output:
4
Time Complexity: O(log2H)
Auxiliary Space: O(1)
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