每日一题_leetcode：day07

# 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:
        def helper(left, right):
            if left > right:
                return None
            # 总是选择中间位置左边的数字作为根节点
            mid = (left + right) // 2
            root = TreeNode(nums[mid])
            root.left = helper(left, mid - 1)
            root.right = helper(mid + 1, right)
            return root
        return helper(0, len(nums) - 1)

还是简单题，最后在归纳吧，这次是递归的简单题，找到根节点之后就很好做。补充一哈：搜索二叉树中序遍历是升序(选择会考)