Leetcode 108 Convert Sorted Array to Binary Search Tree 将一个有序数组变成BST

题目描述：
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    给出一个有序的数组，将这个数组根据BST的定义，改成一棵二叉搜索树，即一个节点的左面的节点都是小于他的，右面的都是大于他的。
思路：
    1.根据定义，有序数组恰好是BST的中序遍历数组，将数组[lo,hi)从mi切分成3片，[lo,mi)  mi  [mi+1,hi)  （数组总是使用[a,b)表示，），分别是左孩子，根节点，右孩子，BST中，处处遵循这样的结构。
    2.时间复杂度O（n）

代码：/**
* 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* sortedArrayToBST(vector<int>& nums)
{
int n=nums.size(),mi=n>>1;//lo=0,hi=n,mi=(lo+hi)/2
if(n == 0) return NULL;
TreeNode* root = new TreeNode(nums[mi]);

vector<int> lnums(nums.begin(), nums.begin()+mi);
vector<int> rnums(nums.begin()+mi+1, nums.end());

root->left = sortedArrayToBST(lnums);
root->right = sortedArrayToBST(rnums);

return root;
}
};