题意：
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：
n = 0 时，空树，只有1棵。n = 1 时，只有1种可能，也是 1。
n >= 2 时，对 12....n，分别以 i 为根节点，那末左侧有 i⑴ 个节点，右侧有 n-i⑴ 个节点，所以
f[n] += f[k⑴]*f[n-k⑴], k = 1,2,....,n⑴
代码：
C++：
class Solution {
public:
int numTrees(int n) {
int *cnt = new int[n+1];
memset(cnt,0,(n+1)*sizeof(int));
cnt[0] = 1;
cnt[1] = 1;
for(int i = 2;i <= n;i++)
for(int j = 0;j < i;++j)
cnt[i] += cnt[j]*cnt[i-j⑴];
int sum = cnt[n];
delete []cnt;
return sum;
}
};
Python：
class Solution:
# @return an integer
def numTrees(self, n):
f = [0 for x in range(0,n+1)]
f[0] = 1
f[1] = 1
for i in range(2,n+1):
for j in range(0,i):
f[i] += f[j]*f[i-j⑴]
return f[n]