# Total number of possible Binary Search Trees using Catalan Number

Given an integer N, the task is to count the number of possible Binary Search Trees with N keys.

Examples:

Input: N = 2
Output: 2
For N = 2, there are 2 unique BSTs
1               2
\            /
2         1

Input: N = 9
Output: 4862


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The number of binary search trees that will be formed with N keys can be calculated by simply evaluating the corresponding number in Catalan Number series.
First few Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …

Catalan numbers satisfy the following recursive formula: Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach  #include  using namespace std;     // Function to return the count  // of unique BSTs with n keys  int uniqueBSTs(int n)  {      int n1, n2, sum = 0;         // Base cases      if (n == 1 || n == 0)          return 1;         // Find the nth Catalan number      for (int i = 1; i <= n; i++) {             // Recursive calls          n1 = uniqueBSTs(i - 1);          n2 = uniqueBSTs(n - i);          sum += n1 * n2;      }         // Return the nth Catalan number      return sum;  }     // Driver code  int main()  {      int n = 2;         cout << uniqueBSTs(n);         return 0;  }

## Java

 // Java implementation of the approach  import java.io.*;     class GFG   {         // Function to return the count  // of unique BSTs with n keys  static int uniqueBSTs(int n)  {      int n1, n2, sum = 0;         // Base cases      if (n == 1 || n == 0)          return 1;         // Find the nth Catalan number      for (int i = 1; i <= n; i++)       {             // Recursive calls          n1 = uniqueBSTs(i - 1);          n2 = uniqueBSTs(n - i);          sum += n1 * n2;      }         // Return the nth Catalan number      return sum;  }     // Driver code  public static void main (String[] args)  {         int n = 2;      System.out.println (uniqueBSTs(n));     }  }     // This code is contributed by jit_t.

## Python3

 # Python3 implementation of the approach     # Function to return the count  # of unique BSTs with n keys  def uniqueBSTs(n):         n1, n2, sum = 0, 0, 0        # Base cases      if (n == 1 or n == 0):          return 1        # Find the nth Catalan number      for i in range(1, n + 1):             # Recursive calls          n1 = uniqueBSTs(i - 1)          n2 = uniqueBSTs(n - i)          sum += n1 * n2         # Return the nth Catalan number      return sum    # Driver code  n = 2    print(uniqueBSTs(n))     # This code is contributed by Mohit Kumar

## C#

 // C# implementation of the approach  using System;     class GFG  {         // Function to return the count  // of unique BSTs with n keys  static int uniqueBSTs(int n)  {      int n1, n2, sum = 0;         // Base cases      if (n == 1 || n == 0)          return 1;         // Find the nth Catalan number      for (int i = 1; i <= n; i++)       {             // Recursive calls          n1 = uniqueBSTs(i - 1);          n2 = uniqueBSTs(n - i);          sum += n1 * n2;      }         // Return the nth Catalan number      return sum;  }     // Driver code  static public void Main ()  {                 int n = 2;      Console.WriteLine(uniqueBSTs(n));     }  }     // This code is contributed by ajit.

Output:

2


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Improved By : jit_t, mohit kumar 29