有序数组转换为平衡二叉搜索树
时间:2020-07-03 12:23:57
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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
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