108. Convert Sorted Array to Binary Search Tree
# Easy
# 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:
# edge case
if len(nums) == 0:
return None
root = TreeNode(nums[0])
if len(nums) == 1:
return root
# regular case
root.val = nums[len(nums)//2]
root.left = self.sortedArrayToBST(nums[:len(nums)//2])
root.right = self.sortedArrayToBST(nums[len(nums)//2+1:])
return root
Time complexity = O(n) , space complexity = O(1) because only need to record the root.