【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);
}
};