https://leetcode.com/problems/unique-binary-search-trees/#/description

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

package go.jacob.day712;

import org.junit.Test;

/**
 * 96. Unique Binary Search Trees
 * 
 * @author Jacob
 *
 */
public class Demo1 {
	/*
	 * 只要確定根節點，那麼左右子樹節點個數就能確定
	 * 
	 * 解析：First note that dp[k] represents the number of BST trees built from
	 * 1....k;
	 * 
	 * Then assume we have the number of the first 4 trees: dp[1] = 1 ,dp[2] =2
	 * ,dp[3] = 5, dp[4] =14 , how do we get dp[5] based on these four numbers
	 * is the core problem here.
	 * 
	 * The essential process is: to build a tree, we need to pick a root node,
	 * then we need to know how many possible left sub trees and right sub trees
	 * can be held under that node, finally multiply them.
	 * 
	 * To build a tree contains {1,2,3,4,5}. First we pick 1 as root, for the
	 * left sub tree, there are none; for the right sub tree, we need count how
	 * many possible trees are there constructed from {2,3,4,5}, apparently it's
	 * the same number as {1,2,3,4}. So the total number of trees under "1"
	 * picked as root is dp[0] * dp[4] = 14. (assume dp[0] =1). Similarly, root
	 * 2 has dp[1]*dp[3] = 5 trees. root 3 has dp[2]*dp[2] = 4, root 4 has
	 * dp[3]*dp[1]= 5 and root 5 has dp[0]*dp[4] = 14. Finally sum the up and
	 * it's done.
	 * 
	 * Now, we may have a better understanding of the dp[k], which essentially
	 * represents the number of BST trees with k consecutive nodes. It is used
	 * as database when we need to know how many left sub trees are possible for
	 * k nodes when picking (k+1) as root.
	 */
	public int numTrees(int n) {
		if (n <= 0)
			return 0;
		int[] res = new int[n + 1];
		res[0] = 1;
		res[1] = 1;
		for (int i = 2; i <= n; i++) {
			for (int j = 1; j <= i; j++) {
				res[i] += res[j - 1] * res[i - j];
			}
		}

		return res[n];
	}
}