# 108. Convert Sorted Array to Binary Search Tree

**<font color=red>难度:Easy<font>**

## 刷题内容
> 原题连接

* https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

> 内容描述

```
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
```

> 思路1
******- 时间复杂度: O(n)******- 空间复杂度: O(1)******

由于是给定的是已排序好的数组，平衡二叉树要求左子树与右字数的差的绝对值不能大于1。因此只需每次将数组对半，其中一半作为左子树，另一半作为右子树即可。这里用assign进行vecrtor赋值。

```cpp
/**
 * 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) {
        if(!nums.size())
            return nullptr;
        TreeNode* root = new TreeNode(nums[nums.size() / 2]);
        vector<int> v1,v2;
        if(nums.size() / 2 >= 0)
            v1.assign(nums.begin(),nums.begin() + nums.size() / 2);
        root ->left = sortedArrayToBST(v1);
        if(nums.size() / 2 + 1 < nums.size())
            v2.assign(nums.begin() + nums.size() / 2 + 1,nums.end());
        root ->right = sortedArrayToBST(v2);
        return root;
    }
};
```