LeetCode OJ:Unique Binary Search Trees

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

算法思想：

1、dp问题，dp[i]表示1到iBST个数，则有

对于新加入一个元素的个数其实也就是将1到i依次作为根节点，比如对于1节点，个数也就是1左边的个数假设为dp[0]乘以1右边的个数dp[i-1]。

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

2、递归

class Solution {
public:
int numTrees(int n) {
if(n==0)return 1;
int sum=0;
for(int i=1;i<=n;i++)
sum+=numTrees(i-1)*numTrees(n-i);
return sum;
}
};

3、根据dp[i]与dp[i-1]之间的关系，也可以把它理解成找规律，很直接

class Solution {
public:
int numTrees(int n) {
int dp = 1;
for (int i = 2; i <= n; i++)
dp = 2*(2*i-1)*dp/(i+1);
return dp;
}
};