題目：

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


思路：
這個問題可以被拆分，若選取值i為根節點，則1,2,...,i - 1為左子樹上的點，i + 1， i + 2, ..., n為右子樹上的點（由搜索二叉樹的性質可得）。而取i為根節點的組合數應為左子樹的種類數*右子樹的種類數。由此可以拆分，得到下邊這種直接叠代的代碼：
 

 1 class Solution(object):
 2     def numTrees(self, n):
 3         """
 4         :type n: int
 5         :rtype: int
 6         """
 7         if n < 1:
 8             return 0
 9         return self.count_tree(1, n)
10 
11     def count_tree(self, left, right):
12         if left >= right:
 
13             return 1
14         res = 0
15         for i in range(left, right + 1):
16             res += self.count_tree(left, i - 1) * self.count_tree(i + 1, right)
17         return res

提交後顯示運行時間超時。原來是代碼中有太多重復叠代，如同求解斐波那契數列時的直接叠代解法。因此使用從底向上的解法：

class Solution(object):
    def numTrees(self, n):
        
 
"""
        :type n: int
        :rtype: int
        """
        if n < 1:
            return 0
        trees = [0 for _ in range(n + 1)]
        trees[0], trees[1] = 1, 1
        for i in range(2, n + 1):
            for j in range(1, i + 1):
                trees[i] += trees[j - 1] * trees[i - j]
        return trees[n]

順利通過~

LeetCode96. Unique Binary Search Trees