[Coding Made Simple] Count number of binary search tree possible given n keys

Given n which is total number of keys in BST, how many BSTs can be formed with n keys.

 

Solution 1. Recursion

public class TotalBst {
    public int getTotalBst(int n) {
        if(n <= 1) {
            return 1;
        }
        int cnt = 0;
        for(int rootIdx = 0; rootIdx < n; rootIdx++) {
            cnt += getTotalBst(rootIdx) * getTotalBst(n - 1 - rootIdx);
        }
        return cnt;
    }
}

 

Solution 2. Dynamic Programming

public int getTotalBst(int n) {
    int[] T = new int[n + 1];
    T[0] = 1;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j < i; j++) {
            T[i] += T[j] * T[i - 1 - j];
        }
    }
    return T[n];
}