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 2^H.

Therefore, from the above observations, print the value of 2^H 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) nodes
int CountBST(int H)
{

    return power(2, H);
}

// Driver Code
int main()
{
    int H = 2;
    cout << CountBST(H);

    return 0;
}

Java

// Java program for the above approach
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) nodes
static long CountBST(int H)
{
    return power(2, H);
}

// Driver code
public 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) nodes
def CountBST(H):

    return power(2, H)

# Driver Code
H = 2

print(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) nodes
static long CountBST(int H)
{

    return power(2, H);
}

// Driver code
static void Main()
{
    int H = 2;

    Console.Write(CountBST(H));
}
}

// This code is contributed by abhinavjain194

Javascript

<script>
// Javascript program for the above approach

var 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) nodes
function CountBST(H)
{

    return power(2, H);
}

// Driver Code
var H = 2;
document.write( CountBST(H));

</script>

Output:

4

Time Complexity: O(log₂H)
Auxiliary Space: O(1)