LeetCode Unique Binary Search Trees

 1 class Solution {
 2 public:
 3     int numTrees(int n) {
 4         return dfs(1, n);
 5     }
 6     
 7     int dfs(int start, int end) {
 8         if (start >= end) return 1;
 9         int count = 0;
10         // choose different number from [start, end] as the root
11         for (int i = start; i <= end; i++) {
12             // number of left tree cases * number of right tree cases
13             count += dfs(start, i - 1) * dfs(i + 1, end); 
14         }
15         return count;
16     }
17 };

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[n];
    }
};