leetcode 【 Convert Sorted List to Binary Search Tree 】python 实现

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

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a list node
    # @return a tree node
    def sortedListToBST(self, head):
        # special case frist
        if head is None:
            return None
        if head.next is None:
            return TreeNode(head.val)
        # slow point & fast point trick to divide the list    
        slow = ListNode(0)
        fast = ListNode(0)
        slow.next = head
        fast.next = head
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        left = head
        right = slow.next.next
        root = TreeNode(slow.next.val)
        slow.next.next = None # cut the connection bewteen right child tree and root TreeNode
        slow.next = None # cut the connection between left child tree and root TreeNode
        root.left = self.sortedListToBST(left)
        root.right = self.sortedListToBST(right)
        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。

http://blog.csdn.net/nandawys/article/details/9125233