108. Convert Sorted Array to Binary Search Tree

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

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

因为BST是有序的，比当前节点小的都放在左子树，比当前节点大的都放在右子树，然后要求是balanced，左右子树的深度不能相差超过1，所以考虑用二分的方法，每次找vector中间的数作为根，然后把左半边的vector作为左子树，继续找中间的数；右半边的vector作为右子树，继续找中间的数。。。依次类推，构建出一棵树。

大体思路是清晰的，但是细节还是有很多问题。

1 使用递归，构建当前子树然后返回根

2 mid = (low + high + 1) / 2, 一定要记得+1，不然到最后的边界会有问题

3 递归函数的参数如何选择，需要用mid - 1代替high， mid+1 代替low，不然会有节点重复的现象

4 什么时候终止递归，这个很重要，每次想的都不是太清楚。这道题是当mid == low时直接把左子树设为NULL，而不继续调用递归函数；当mid == high 时，把右子树设为NULL，而不继续调用递归函数。

用例子 [-10, -3, 0, 5, 9] 做一个状态转移，分析一下边界问题，和终止条件：

代码如下：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* builtTree(vector<int> nums, int low, int high) {        int mid = (high + low + 1) / 2; //remember to +1 or mid will be wrong        TreeNode* root = new TreeNode(nums[mid]);        if (mid == low) root->left = NULL; // terminal condition        else root->left = builtTree(nums, low, mid - 1); // mid - 1 instead of mid or node will repeat        if (mid == high) root->right = NULL; // terminal condition        else root->right = builtTree(nums, mid + 1, high); // mid + 1 same as mid - 1        return root;    }    TreeNode* sortedArrayToBST(vector<int>& nums) {        if (nums.size() == 0) return NULL;        if (nums.size() == 1) {            TreeNode* root = new TreeNode(nums[0]);            return root;        }        return builtTree(nums, 0, nums.size() - 1);    }};