Unique Binary Search Trees

其实和大小没啥太大关系,主要因素是元素个数,也就是说,1,2 和 3,4 可能构成的排列组合个数是一样的,大概的思路就是这样的:

式子出来了,那代码也就出来了:

int numTrees(int n) {
    vector<int> treeNumList(n + 1);
    function<int(int)> getTreeNum;
    getTreeNum = [&](int num)
    {
        if (num == 0 || num == 1){
            return 1;
        }
        int result = 0;
        if ((result = treeNumList[num]) != 0){
            return result;
        }
        bool const isOdd = (num % 2 != 0);
        const int k = isOdd? (num - 1) / 2 : num / 2;
        for (int i = 0; i != k; ++i){
            result += 2 * getTreeNum(i) * getTreeNum(num - i - 1);
        }
        if (isOdd){
            result += getTreeNum(k) * getTreeNum(k);
        }
        treeNumList[num] = result;
        return result;
    };
    return getTreeNum(n);
}
原文地址:https://www.cnblogs.com/wuOverflow/p/5046950.html