【LeetCode】Unique Binary Search Trees，leetcodebinary

【LeetCode】Unique Binary Search Trees，leetcodebinary
     題意：

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 時，空樹，只有一棵。n = 1 時，只有一種可能，也是 1。

    n >= 2 時，對 12....n，分別以 i 為根節點，那麼左邊有 i-1 個節點，右邊有 n-i-1 個節點，所以

        f[n] += f[k-1]*f[n-k-1], k = 1,2,....,n-1

    代碼：

    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-j-1];        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-j-1]                return f[n]