108. 将有序数组转换为二叉搜索树

1. 题目链接

2. 解题思路：这里面有一个构造「平衡二叉树」，似乎很难。实际上，我们每次构造时都拿中点划分，就能得到平衡的。

   • 具体来说process(nums, L, R)在nums[L, R]上构造平衡搜索二叉树，我们以中点mid=(R+L)/2是头，然后左节点process(nums, L, mid - 1)，右节点process(nums, mid + 1, R)

3. 代码

   class Solution:
   
       def process(self, nums: List[int], L: int, R: int) -> Optional[TreeNode]:
           if L > R:
               return None
           if L == R:
               return TreeNode(nums[L])
           mid = int((R + L) / 2)
           head = TreeNode(nums[mid])
           left_head = self.process(nums, L, mid - 1)
           right_head = self.process(nums, mid + 1, R)
           head.left = left_head
           head.right = right_head
           return head
   
   
       def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
           return self.process(nums, 0, len(nums) -1)