109. Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树
【LeetCode】109. Convert Sorted List to Binary Search Tree 解题报告(Python)
标签(空格分隔): LeetCode
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/
题目地址:https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/
题目描述:
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
题目大意
把一个单链表转成二叉平衡树。二叉平衡树是左右子树的高度差最多为1.
解题方法
这个题目和108. Convert Sorted Array to Binary Search Tree基本一模一样啊!所以毫无疑问地,我把链表转成了数组,然后再用108题的做法就过了。
而且这个题神奇的地方在于,使用python2超时,但是使用Python3就通过了。以后还是尽量用python3比较好吧。。
哦对,python3要用地板除。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sortedListToBST(self, head):
"""
:type head: ListNode
:rtype: TreeNode
"""
nums = []
while head:
nums.append(head.val)
head = head.next
return self.helper(nums)
def helper(self, nums):
if not nums: return None
_len = len(nums)
mid = _len // 2
root = TreeNode(nums[mid])
root.left = self.helper(nums[:mid])
root.right = self.helper(nums[mid+1:])
return root
日期
2018 年 6 月 23 日 ———— 美好的周末要从刷题开始