LeetCode_Convert Sorted Array to Binary Search Tree（Java实现）

题目描述：

给出一个升序排序的数组，将其转化为平衡二叉搜索树（BST）.

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

对于此问题，将高度平衡的二叉树定义为一个二叉树，其中每个节点的两个子树的深度相差不超过1。

例：

Given the sorted array: [-10,-3,0,5,9],One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:0/ \-3   9/   /-10  5

主要思想：

把一个有序数组转化为一棵二叉搜索树，二叉搜索树（BST）具有以下性质：

若其左子树存在，则其左子树中每个节点的值都不大于该节点值；
若其右子树存在，则其右子树中每个节点的值都不小于该节点值。

示例：

因此，可以用递归来构建一棵二叉搜索树，每次把数组分为两半，把数组中间的值作为其父节点，然后把数组的左右两部分进行递归，继续构造其左右子树。

实现代码：

/*** Definition for binary tree* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public TreeNode sortedArrayToBST(int[] num) {if ((num == null) || (num.length == 0)) {return null;}return helper(num, 0, num.length - 1);}private TreeNode helper(int[] num, int left, int right) {if (left > right) {return null;}int mid = (left + right + 1) >> 1;TreeNode node = new TreeNode(num[mid]);node.left = helper(num, left, mid - 1);node.right = helper(num, mid + 1, right);return node;}
}

注意：在牛客网上的环境下，mid = (left + right + 1) / 2，如果没有 +1，不能通过所有的测试样例；

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