Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Number of Binary Search Trees of height H consisting of H+1 nodes

  • Last Updated : 19 May, 2021

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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) 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)

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!