Total number of possible Binary Search Trees using Catalan Number

• Difficulty Level : Medium
• Last Updated : 16 Jun, 2022

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 keysint 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 codeint main(){    int n = 2;     // Function call    cout << uniqueBSTs(n);     return 0;}

Java

 // Java implementation of the approachimport 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 coden = 2 # Function callprint(uniqueBSTs(n)) # This code is contributed by Mohit Kumar

C#

 // C# implementation of the approachusing 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

 

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 1â€¦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 1â€¦(i-1) on its left side would lay on the left branch of the root, and similarly the right subsequence (i+1)â€¦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 1â€¦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) + â€¦ + 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) â€¦ 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 keysint 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 codeint main(){    int n = 2;     // Function call    cout << uniqueBSTs(n);     return 0;}

Java

 // Java dynamic programming implementation of the approachimport 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 keysdef 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 codeif __name__ == '__main__':         n = 2     # Function call    print(uniqueBSTs(n)) # This code is contributed by bgangwar59

C#

 // C# dynamic programming implementation// of the approachusing System; class GFG{   // Function to return the count// of unique BSTs with n keysstatic 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 codepublic static void Main(String[] args){    int n = 2;       // Function call    Console.WriteLine(uniqueBSTs(n));}} // This code is contributed by Amit Katiyar

Javascript

 

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  which simplifies to

Similarly Catalan number for (n-1) nodes  =

The formula for n nodes can be rewritten as

= Catalan number for (n-1) nodes*

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 using namespace std; // Function to find number of unique BSTint 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;}

Java

 class GFG{ // Function to find number of unique BSTstatic long 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;} // Driver codepublic static void main(String[] args){    int n = 4;    System.out.print("Number of Unique BST for " + n +                     " nodes is " + numberOfBST(n) + "\n");}} // This code is contributed by shikhasingrajput

Python3

 # Function to find number of unique BSTdef numberOfBST(n):       # For n=1 answer is 1    v = 1    for i in range(2, n + 1):               # using previous answer in v to calculate current catalan number.        v = ((v * (i * 2) * (i * 2 - 1)) / ((i + 1) * (i)))    return int(v) n = 4print("Number of Unique BST for", n, "nodes is", numberOfBST(n)) # This code is contributed by divyesh072019.

C#

 using System;class GFG {         // Function to find number of unique BST    static int numberOfBST(int n)    {               // For n=1 answer is 1        int 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;    }   static void Main() {    int n = 4;    Console.Write("Number of Unique BST for " + n + " nodes is " + numberOfBST(n));  }} // This code is contributed by rameshtravel07.

Javascript

 

Output

Number of Unique BST for 4 nodes is 14

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

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