【LeetCode笔记】Convert Sorted Array to Binary Search Tree 通过有序数列建立二叉搜索树

题目：

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

思路：每一次从有序数列的中间开始提一个出来作为当前的根节点。然后其左边就是左子树，右边就是右子树。然后对左右子树分别递归~

/**
 * 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* build(vector<int>& nums,int start, int end){
        if(start>end)
            return NULL;
        else{
            
            int m = (start+end)/2;
            TreeNode* t = new TreeNode(nums[m]);
            t->left = build(nums,start,m-1);  //注意，这里是m-start-1+start=m-1!!!!因为是基于start开始计数
            t->right = build(nums,m+1,end);
            return t;
        }
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return build(nums,0,nums.size()-1);
    }
};