# \[108]\[简单]\[DFS]\[二分] 将有序数组转换为二叉搜索树

## 题目描述

[108. 将有序数组转换为二叉搜索树](https://leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree/)

将一个按照升序排列的有序数组，转换为一棵高度平衡二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

```
给定有序数组: [-10,-3,0,5,9],

一个可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面这个高度平衡二叉搜索树：

      0
     / \
   -3   9
   /   /
 -10  5
```

## 解题思路

[将有序数组转换为二叉搜索树](https://leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree/solution/jiang-you-xu-shu-zu-zhuan-huan-wei-er-cha-sou-s-33/)

简单来说, **二叉搜索树的中序遍历是升序序列**. 因此可将数组中间的数作为树的根节点, 左右子数组分别构成根节点的左右子树. 而左右子树的构建方法同上, 因此依据DFS的思路, 使用递归逐步得到各级子树.

注意边界条件, 当`left`大于`right`时, 说明此时构成的树为空树.

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        def get_tree(left, right):
            if left > right:
                return None

            mid = (left + right) // 2
            root = TreeNode(nums[mid])
            root.left = get_tree(left, mid - 1)
            root.right = get_tree(mid + 1, right)
            return root
        return get_tree(0, len(nums) - 1)
```

## 相关题目

* 链表版: [\[109\]\[中等\]\[DFS\] 有序链表转换二叉搜索树](/garnet/suan-fa/shuang-zhi-zhen/109-you-xu-lian-biao-zhuan-huan-er-cha-sou-suo-shu.md)


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