LeeCode - Unique Binary Search Trees

题目：

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

思路：

找规律

1. 没有节点，BST （Binary Search Tree）为1个， 即空树

2. 有一个节点，BST为1个，即自身为根节点

3. 有二个节点，BST为2个，1为root，左为空，右为2；2为root，左为1，右为空，其实就是dp[0]*dp[1] + dp[1]*dp[0]

1 2

\ /

2 1

4. 有三个节点

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

可以推出dp[3] = dp[0]*dp[2] + dp[1]*dp[1] + dp[2]*dp[0]

package bst;

public class UniqueBinarySearchTrees {

public int numTrees(int n) {
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
dp[i] += dp[j] * dp[i - 1 - j];
}
}
return dp
;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
UniqueBinarySearchTrees u = new UniqueBinarySearchTrees();
System.out.println(u.numTrees(3));
}

}