【leetcode】96. Unique Binary Search Trees

题目如下：

Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?

Example:

Input: 3
Output: 5
Explanation:
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

解题思路：一开始我觉得递归就行了，根节点可以填入的数为[1,2,3....n]，如果选择该根节点值为2，那么其左边子树可以选择的就是[1]，而右边子树可以选择的就是[3...n]，所以根节点为2可以构成的树的总量就是左边子树的种数*右边子树的总数，依次累计根节点填入1~n的总数即可。

代码如下：

class Solution(object):
    dic = {}
    def recursive(self,nl):
        if (nl[0],nl[-1]) in self.dic:
            return self.dic[(nl[0],nl[-1])]
        if len(nl) == 1:
            self.dic[(nl[0], nl[-1])] = 1
            return 1
        count = 0
        for i in range(len(nl)):
            l1 = nl[:i] if i > 0 else []
            l2 = nl[i+1:]
            if len(l1) + len(l2) == 0:
                count += 0
            elif len(l1) == 0:
                count += self.recursive(l2)
            elif len(l2) == 0:
                count += self.recursive(l1)
            else:
                count += (self.recursive(l1) * self.recursive(l2) )
        self.dic[(nl[0], nl[-1])] = count
        return count

    def numTrees(self, n):
        """
        :type n: int
        :rtype: int
        """
        self.dic = {}
        return self.recursive(range(1,n+1))