Total number of possible Binary Search Trees using Catalan Number

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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <iostream>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 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 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# 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.
<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>
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++
Java
Python3
C#
Javascript
// C++ dynamic programming implementation of the approach#include <iostream>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 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 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# 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
<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>
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++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>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;}
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
# 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.
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.
<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>
Number of Unique BST for 4 nodes is 14
Time Complexity: O(n)Auxiliary Space: O(1). 
jit_t
mohit kumar 29
AkashAryan
bgangwar59
avanitrachhadiya2155
divyeshrabadiya07
vaibhavrabadiya117
amit143katiyar
rahul mishra
mukesh07
divyesh072019
rameshtravel07
shikhasingrajput
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