108. 将有序数组转换为二叉搜索树

Easy

思路

思路其实很简单,明显的递归题

以上,尝试写一下代码,AC!

代码

python3

class Solution:
    def build(self, nums):
        if len(nums) is 0:
            return None
        mid = len(nums) // 2
        m = nums[mid]
        left = nums[:mid]
        right = nums[mid+1:]
        node = TreeNode(m)
        node.left = self.build(left)
        node.right = self.build(right)
        return node 

    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        return self.build(nums)