【LeetCode & 剑指offer刷题】树题17：Convert Sorted Array to Binary Search Tree
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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 everynode 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
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
方法：分治法（分成处理左右子序列的问题，与快速排序写法类似）
二叉查找树，根结点总是大于左子树所有结点值，小于右子树所有结点值，中序遍历序列为升序排列
分析：转化成一个height-balanced的二叉查找树，所以数组中点就是二叉查找树的根结点
写递归思路：先写递归的主体，再写起点输入，再写递归的出口return语句（尤其注意递归的子函数的出口）
*/
class Solution
{
public:
TreeNode* sortedArrayToBST(vector<int>& nums)
{
return buildTree(nums,0,nums.size()-1); //注意索引从0开始
}
TreeNode* buildTree(vector<int>& nums, int start, int end)
{
if(start > end) return NULL; //递归子函数的出口（不能取等号，因为单个元素也要分配空间）
int mid = (start + end + 1)/2; //+1是为了向上取整，由于是升序排列，当(start+end)/2不为整数时，取较大的那个数作为根
TreeNode* root = new TreeNode(nums[mid]); //构建根结点
root->left = buildTree(nums, start, mid-1); //构建左子树
root->right = buildTree(nums, mid+1, end); //构建右子树
return root; //递归原始母函数的出口，返回最顶层的根结点指针
}
};
