109. Convert Sorted List to Binary Search Tree 有序链表转换二叉搜索树

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.

Forthis 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],

Onepossible 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 Treeopen in new window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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

DDKK.COM 弟弟快看-教程，程序员编程资料站，版权归原作者所有

本文经作者：负雪明烛 授权发布，任何组织或个人未经作者授权不得转发