96. Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

Solution：DP

思路：

G(n): the number of unique BST for a sequence of length n.
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.

G(0)=1, G(1)=1. 
G(n) = F(1, n) + F(2, n) + ... + F(n, n). 
F(i, n) = G(i-1) * G(n-i)   1 <= i <= n 
So, 
G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0) 

Time Complexity: O(N^2) Space Complexity: O(N)

Solution Code：

class Solution {
    public int numTrees(int n) {
        int [] G = new int[n+1];
        G[0] = G[1] = 1;

        for(int i=2; i<=n; ++i) {
            for(int j=1; j<=i; ++j) {
                G[i] += G[j-1] * G[i-j];
            }
        }
        return G[n];
    }
}