96. 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

给定一个序列1~n，为了构造二叉树，我们可以使用1~n中的第 i 个节点作为根节点，则根据二叉排序树的性质，1~n-1必在左子树，n+1~n在右子树；

f(0)= 1；
f(1) = 1；
f(2) = f(1) * f(0) + f(0) * f(1)；
以第 i 个节点作为根节点有f(i - 1) * f( n - 1 - i)中二叉排序树的可能；
则f(n) = f(0) * f(n - 1) + f(1) * f(n-2)+…+f(n-1) * f(0)

思路1：用一个二重循环

int numTrees(int n) {    if (n == 0)return 0;    if (n == 1)return 1;    vector<int> res(n + 1, 0);    res[0] = 1, res[1] = 1;    for (int i = 2; i <= n; i++){        for (int j = 1; j <= i; j++){            res[i] += res[j - 1] * res[i - j];        }    }    return res[n];}

思路2：递归

int numTrees(int n) {    if (n == 0 || n == 1)return 1;    int res = 0;    for (int i = 0; i < n; i++){        res += numTrees(i) * numTrees(n - i - 1);    }    return res;}

思路3：数学公式直接求解

int numTrees(int n) {    if (n == 0)return 0;    if (n == 1)return 1;    vector<long>res(n + 1, 0);    res[0] = 1, res[1] = 1;    for (int i = 1; i < n; i++){        res[i + 1] = 2 * (2 * i + 1) *res[i] / (i + 2);    }    return res[n];}