有序数组转换为平衡二叉搜索树

Posted 2020-12-14 gugu-da

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

• 时间复杂度：O(n) n为数组长度
• 空间复杂度：O(logn)

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 helper(left,right):
            if left>right:
                return None
            mid = (left+right+1)//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)

递归思想