108将有序数组转化为二查搜索树

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
from typing import List
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        # 如果列表为空的话，就返回空节点
        if nums == [] : return
        # 取列表中间的那个值。当做根节点
        mid = len(nums) // 2
        mid_value = nums[mid]
        root = TreeNode(mid_value)
        # 列表左边的值为根节点的左儿子，
        root.left = self.sortedArrayToBST(nums[:mid])
        # 右边的值为根节点的右儿子
        root.right = self.sortedArrayToBST(nums[mid + 1 :])
        return root