递归，每次取有序数组中间的数创建当前节点，同时有序数组前半用于构建当前节点的左子树，后半用于构建当前节点的右子树

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:

    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        root = TreeNode()
        self.sortedArrayToBSTHelper(root,nums)
        return root
    
    def sortedArrayToBSTHelper(self,root:TreeNode,nums:List[int]):
        L = len(nums)
        root.val = nums[L//2]
        if(L>2):
            root.left  = TreeNode()
            root.right = TreeNode()
            self.sortedArrayToBSTHelper(root.left,nums[:L//2]) #构建左子树
            self.sortedArrayToBSTHelper(root.right,nums[L//2+1:]) #构建右子树
        elif(L==2):
            root.left= TreeNode(val=nums[0])

        else:
            return 

递归yyds，希望这次学了构建搜索树别再忘了…