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Total number of possible Binary Search Trees using Catalan Number

  • Difficulty Level : Medium
  • Last Updated : 09 Nov, 2021

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++




// 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;
}

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.

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the count
    // of unique BSTs with n keys
    function uniqueBSTs(n)
    {
        let n1, n2, sum = 0;
  
        // Base cases
        if (n == 1 || n == 0)
            return 1;
  
        // Find the nth Catalan number
        for (let 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;
    }
     
    let n = 2;
        
    // Function call
    document.write(uniqueBSTs(n));
 
</script>
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)        
        /    |          \
    G(0)G(2)  G(1)G(1)  G(2)G(0)                               
    /     \
G(0)G(1)  G(1)G(0) // base case                        

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 is 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++




// 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;
}

Java




// Java dynamic programming implementation of the approach
import java.io.*;
import java.util.*;
class GFG
{
   
    // Function to return the count
    // of unique BSTs with n keys
    static int uniqueBSTs(int n)
    {
       
        // construct a dp array to store the
        // subsequent results
        int[] dparray = new int[n + 1];
        Arrays.fill(dparray, 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
    public static void main (String[] args)
    {
        int n = 2;
       
        // Function call
        System.out.println(uniqueBSTs(n));
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 dynamic programming
# implementation of the approach
 
# Function to return the count
# of unique BSTs with n keys
def uniqueBSTs(n):
   
    # Construct a dp array to store the
    # subsequent results
    dparray = [0 for i in range(n + 1)]
 
    # There is only one combination to
    # construct a BST out of a sequence of
    dparray[0] = 1
    dparray[1] = 1
     
    # length 1 (only a root) or 0 (empty tree).
    for i in range(2, n + 1, 1):
         
        # Choosing every value as root
        for k in range(1, i + 1, 1):
            dparray[i] += (dparray[k - 1] *
                           dparray[i - k])
                            
    return dparray[n]
 
# Driver code
if __name__ == '__main__':
     
    n = 2
 
    # Function call
    print(uniqueBSTs(n))
 
# This code is contributed by bgangwar59

C#




// C# dynamic programming implementation
// of the approach
using System;
 
class GFG{
   
// Function to return the count
// of unique BSTs with n keys
static int uniqueBSTs(int n)
{
     
    // construct a dp array to store the
    // subsequent results
    int[] dparray = new int[n + 1];
   
    // 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
public static void Main(String[] args)
{
    int n = 2;
   
    // Function call
    Console.WriteLine(uniqueBSTs(n));
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
    // Javascript dynamic programming
    // implementation of the approach
     
    // Function to return the count
    // of unique BSTs with n keys
    function uniqueBSTs(n)
    {
        
        // construct a dp array to store the
        // subsequent results
        let dparray = new Array(n + 1);
        dparray.fill(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 (let i = 2; i <= n; ++i)
        {
            
            // choosing every value as root
            for (let k = 1; k <= i; ++k)
            {
                dparray[i] += dparray[k - 1] * dparray[i - k];
            }
        }
        return dparray[n];
    }
     
    let n = 2;
        
    // Function call
    document.write(uniqueBSTs(n));
 
</script>
Output



2

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

In this post, we will discuss an O(n) and an O(1) space solution based on Dynamic Programming.

 We know that the formula for Catalan number for a variable n  is \frac{1}{(n+1)}*\binom{2n}{n}    which simplifies to \frac{2n!}{n!(n+1)!}

Similarily Catalan number for (n-1) nodes  = \frac{2(n-1)!}{n!(n-1)!}

The formula for n nodes can be rewritten as \frac{2(n-1)!}{n!*(n-1)!}*\frac{2n*(2n-1)}{(n+1)*n}

                                                        = Catalan number for (n-1) nodes* \frac{2n*(2n-1)}{(n+1)*n}

So for every iteration for ‘i’ going from 1 to n we will store catalan number for ‘i-1’ nodes and compute for ith node.

Below is the implementation for the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to find number of unique BST
int numberOfBST(int n)
{
    // For n=1 answer is 1
    long v = 1;
    for (int i = 2; i <= n; i++) {
        // using previous answer in v to calculate current
        // catalan number.
        v = ((v * (i * 2) * (i * 2 - 1)) / ((i + 1) * (i)));
    }
    return v;
}
 
int main()
{
    int n = 4;
    cout << "Number of Unique BST for " << n << " nodes is "
         << numberOfBST(n) << endl;
    return 0;
}

Javascript




<script>
    // Function to find number of unique BST
    function numberOfBST(n)
    {
        // For n=1 answer is 1
        let v = 1;
        for (let i = 2; i <= n; i++) {
            // using previous answer in v to calculate current
            // catalan number.
            v = ((v * (i * 2) * (i * 2 - 1)) / ((i + 1) * (i)));
        }
        return v;
    }
     
    let n = 4;
    document.write("Number of Unique BST for " + n + " nodes is "
         + numberOfBST(n));
    
   // This code is contributed by mukesh07.
</script>
Output
Number of Unique BST for 4 nodes is 14

Time Complexity: O(n)
Auxiliary Space: O(1). 




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