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

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: 

C_0=1 \ and \ C_n_+_1=\sum_{i=0}^{n}C_iC_n_-_i \ for \ n\geq 0;

Below is the implementation of the above approach:



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <iostream>
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;
 
    // Function call
    cout << uniqueBSTs(n);
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
         
        // Function call
        System.out.println(uniqueBSTs(n));
    }
}
 
// This code is contributed by jit_t.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
 
# Function call
print(uniqueBSTs(n))
 
# This code is contributed by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
       
        // Function call
        Console.WriteLine(uniqueBSTs(n));
    }
}
 
// This code is contributed by ajit.

chevron_right


Output

2

The problem can be solved in a dynamic programming way. 
Here is a snippet of how the recurrence tree will proceed: 

                               G(4)
                     /      |        |        \
             G(0)G(3)     G(1)G(2)  G(2)G(1)   G(3)G(0)        // 4
            /    |    \
    G(0)G(2)  G(1)G(1)  G(2)G(0)                               // 4 x 3
    /     \
G(0)G(1)  G(1)G(0) // base case                                // 4 x 3 x 2

Note: Without memoization, the time complexity is upper bounded by O(N x N!).
Given a sequence 1…n, to construct a Binary Search Tree (BST) out of the sequence, we could enumerate each number i in the sequence, and use the number as the root, naturally, the subsequence 1…(i-1) on its left side would lay on the left branch of the root, and similarly the right subsequence (i+1)…n lay on the right branch of the root. We then can construct the subtree from the subsequence recursively. Through the above approach, we could ensure that the BST that we construct are all unique, since they have unique roots. 

The problem is to calculate the number of unique BST. To do so, we need to define two functions: 

1.G(n): the number of unique BST for a sequence of length n.
2.F(i, n), 1 <= i <= n: The number of unique BST, where the number i is 
the root of BST, and the sequence ranges
from 1 to n. As one can see, G(n) is the actual function we need to calculate in order to 
solve the problem. And G(n) can be derived from F(i, n), which at the end, would recursively 
refer to G(n).
First of all, given the above definitions, we can see that the total number of unique BST G(n),
is the sum of BST F(i) using each number i as a root. i.e.,
G(n) = F(1, n) + F(2, n) + ... + F(n, n).
Given a sequence 1…n, we pick a number i out of the sequence as the root, then the number of 
unique BST with the specified root F(i), is the cartesian product of the number of BST for 
its left and right subtrees.For example, F(2, 4): the number of unique BST tree with number 2 
as its root. To construct an unique BST out of the entire sequence [1, 2, 3, 4] with 2 as the 
root, which is to say, we need to construct an unique BST out of its left subsequence [1] and 
another BST out of the right subsequence [3,4], and then combine them together (i.e. cartesian 
product). F(i, n) = G(i-1) * G(n-i)    1 <= i <= n 
Combining the above two formulas, we obtain the recursive formula for G(n). i.e.

G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0) 

In terms of calculation, we need to start with the lower number, since the value of G(n) 
depends on the values of G(0) … G(n-1). 

Below is the above implementation of the above algorithm:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ dynamic programming implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the count
// of unique BSTs with n keys
int uniqueBSTs(int n)
{
    // construct a dp array to store the
    // subsequent results
    int dparray[n + 1] = { 0 };
 
    // there is only one combination to construct a
    // BST out of a sequence of
    dparray[0] = dparray[1] = 1;
     
    // length 1 (only a root) or 0 (empty tree).
    for (int i = 2; i <= n; ++i)
    {
        // choosing every value as root
        for (int k = 1; k <= i; ++k)
        {
            dparray[i] += dparray[k - 1] * dparray[i - k];
        }
    }
    return dparray[n];
}
 
// Driver code
int main()
{
    int n = 2;
 
    // Function call
    cout << uniqueBSTs(n);
 
    return 0;
}

chevron_right


Output

2

Time Complexity: O(N2) 
Space Complexity: O(N)

competitive-programming-img




My Personal Notes arrow_drop_up

I like to write Technical Articles on Data Structures and Algorithm in my leisure time

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.