108.将有序数组转换为二叉搜索树
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def sortedArrayToBST(self, nums: List[int]) -> TreeNode: def to_bst(nums, start, end): if start > end: return None mid = (start + end) // 2 node = TreeNode(nums[mid]) node.left = to_bst(nums, start, mid-1) node.right = to_bst(nums, mid+1, end) return node return to_bst(nums, 0, len(nums) - 1)