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

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  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++

 // C++ dynamic programming implementation of the approach #include  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 = dparray = 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; }

Output

2

Time Complexity: O(N2)
Space Complexity: O(N) My Personal Notes arrow_drop_up I like to write Technical Articles on Data Structures and Algorithm in my leisure time

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