题目如下：
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
解题思路：题目没有明确要求不允许使用额外的内存，所以最简单的方法是把linked list中每个元素的值存入list，然后套用【leetcode】108. Convert Sorted Array to Binary Search Tree 的解法。
代码如下：
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def recursive(self,node,nums): mid = len(nums)/2 left_num = nums[:mid] if len(left_num) > 0: node.left = TreeNode(left_num[len(left_num)/2]) self.recursive(node.left,left_num) right_num = nums[mid+1:] if len(right_num) > 0: node.right = TreeNode(right_num[len(right_num)/2]) self.recursive(node.right,right_num) def sortedListToBST(self, head): """ :type head: ListNode :rtype: TreeNode """ nums = [] while head != None: nums.append(head.val) head = head.next if len(nums) == 0: return None root = TreeNode(nums[len(nums)/2]) self.recursive(root,nums) return root
【leetcode】109. Convert Sorted List to Binary Search Tree
原文：https://www.cnblogs.com/seyjs/p/10625312.html