问题描述:
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.
\ / / / \ \ / / \ \
解决原理1:
遍历+递归
二叉查找树的根节点可以是1~n中的任何一个数i
根为i的二叉树数量=左子树数量*右子树数量
左子树的根节点取值范围为1~i,右子树的根节点取值范围为i+1~n,i+1~n组成的二叉查找树的数量又与1~n-i的相同
代码1:
class Solution {
int sum;
public:
int numTrees(int n) {
if(n == ) return ;
if(n == ) return ;
for(int i = ; i <= n; i++){
int l = numTrees(i-);
int r = numTrees(n-i);
sum = sum + (l==?:l) * (r==?:r);
}
return sum;
}
};
但是,此方法超时