LeetCode Unique Binary Search Trees

class Solution {
public:
int numTrees(int n) {
return dfs(1, n);
}

int dfs(int start, int end) {
if (start >= end) return 1;
int count = 0;
// choose different number from [start, end] as the root
for (int i = start; i <= end; i++) {
// number of left tree cases * number of right tree cases
count += dfs(start, i - 1) * dfs(i + 1, end);
}
return count;
}
};

想到了就简单，再水一发

第二轮:

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

就是卡特兰数其实，f(n) = f(0) * f(n-1) + f(1) * f(n-2)... + f(n-1) * f(0)，题目中递增序列具体区间没有关系，只要关心其内数字个数即可：

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