题目：
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
代码：oj测试通过 Runtime: 178 ms
1 # Definition for a binary tree node 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 # 8 # Definition for singly-linked list. 9 # class ListNode: 10 # def __init__(self, x): 11 # self.val = x 12 # self.next = None 13 14 class Solution: 15 # @param head, a list node 16 # @return a tree node 17 def sortedListToBST(self, head): 18 # special case frist 19 if head is None: 20 return None 21 if head.next is None: 22 return TreeNode(head.val) 23 # slow point & fast point trick to divide the list 24 slow = ListNode(0) 25 fast = ListNode(0) 26 slow.next = head 27 fast.next = head 28 while fast.next is not None and fast.next.next is not None: 29 slow = slow.next 30 fast = fast.next.next 31 left = head 32 right = slow.next.next 33 root = TreeNode(slow.next.val) 34 slow.next.next = None # cut the connection bewteen right child tree and root TreeNode 35 slow.next = None # cut the connection between left child tree and root TreeNode 36 root.left = self.sortedListToBST(left) 37 root.right = self.sortedListToBST(right) 38 return root
思路：
binary search tree 是什么先搞清楚
由于是有序链表，所以可以采用递归的思路，自顶向下建树。
1. 每次将链表的中间节点提出来；链表中间节点之前的部分作为左子树继续递归；链表中间节点之后的部分作为右子树继续递归。
2. 停止递归调用的条件是传递过去的head为空(某叶子节点为空)或者只有一个ListNode（到某叶子节点了）。
找链表中间节点的时候利用快慢指针的技巧：注意，因为前面的special case已经将传进来为空链表和长度为1的链表都处理了，所以快慢指针的时候需要判断一下从最短长度为2的链表的处理逻辑。之前的代码在while循环中只判断了fast.next.next is not None就忽略了链表长度为2的case，因此补上了一个fast.next is not None的case，修改过一次就AC了。
网上还有一种思路，只需要走一次链表就可以完成转换，利用的是自底向上建树。下面这个日志中有说明，留着以后去看看。