154-108. 将有序数组转换为二叉搜索树

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。(一开始没思路,然后看了一下官方题解,稀里糊涂就做出来了,我也蒙了)
# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        start = 0
        end = len(nums)
        mid = (start + end) // 2
        if start >= end:
            return
        root = TreeNode(nums[mid])
        left = self.sortedArrayToBST(nums[0: mid])
        right = self.sortedArrayToBST(nums[mid + 1: end])
        if left:
            root.left = left
        if right:
            root.right = right
        return root


if __name__ == '__main__':
    s1 = Solution()
    nums = [-10, -3, 0, 5, 9]
    root = s1.sortedArrayToBST(nums)
    print("gfsdg")