Unique Binary Search Trees

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

其实很容易想到，对于一个根结点，所有情况是左子树个数与右子树个数的乘积，而对于n个结点，所有情况为对于每一个结点作为根时的情况的和。一开始直接写了一段递归，明知道过不了大集合还是试了一下，结果小集合也没有过，所以干脆bottom-up吧。代码：

int numTrees(int n) {
  // IMPORTANT: Please reset any member data you declared, as
  // the same Solution instance will be reused for each test case.
  int* cache = (int*)malloc(sizeof(int)*(n+1));
  int i,j,sum;
  cache[0]=1;
  cache[1]=1;
  //printf("%d\r\n",cache[0]);
  for(i=2;i<=n;i++){
    sum=0;
    for(j=0;j<i;j++)
      sum+=cache[j]*cache[i-1-j];
    cache[i]=sum;
    //printf("%d\r\n",cache[i]);    
  }
  int result = cache[n];
  free(cache);
  return result;
}