LeetCode 96. Unique Binary Search Trees

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
二分查找树的定义是，左子树节点均小于root，右子树节点均大于root~~~
所以可以用递推的方法，把v[i]表示i个数能够构成的二叉搜索树的个数~~
初始化边界值是 v[0]=1,v[1]=1,v[2]=2~~~
当i>=3的时候，若以j为root结点，v[j-1]等于root结点左边的j-1个结点能构成的BST个数~~
v[i-j]等于root结点右边i-j个结点能构成的BST个数~//j+1~i的种数和0~i-j的种数一样。。所以就是v[i-j]~
所以v[j-1] * v[i-j]等于以j为root结点能构成的BST种数~~~
j可以取1~i中的任意一个值，把这些所有计算出来的总数相加就是v[i]的值~~~~
所以 for(int j = 1; j <= i; j++) {
v[i] += v[j-1] * v[i-j];
}

最后返回的值是v[n]的值，表示1~n能组成的BST的个数~~~~

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

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