Problem: 96. 不同的二叉搜索树

题目描述

思路

一个数字做根节点的话可能的结果为：其左边数字做子树的组合数字乘以其右边数字做子树的个数之积

1.创建备忘录memo；
2.递归分别求取当前数字左边和右边数字做子树的数量（注意下面代码当左边界值大于有边界值时应当反回1）

复杂度

时间复杂度:

$O(n)$；其中$n$是二叉树节点的个数

空间复杂度:

$O(height)$;其中$height$是二叉树的高度

Code

class Solution {int[][] memo;/*** Unique Binary Search Trees** @param n Given number* @return int*/public int numTrees(int n) {memo = new int[n + 1][n + 1];return count(1, n);}/*** Unique Binary Search Trees(Implementation function)** @param low  Left boundary* @param high Right boundary* @return int*/private int count(int low, int high) {if (low > high) {return 1;}//Check the memeif (memo[low][high] != 0) {return memo[low][high];}int res = 0;for (int mid = low; mid <= high; ++mid) {int left = count(low, mid - 1);int right = count(mid + 1, high);res += left * right;}memo[low][high] = res;return res;}
}