109. Convert Sorted List to Bina

将链表转化为平衡二叉查找树，和数组转化为平衡二叉查找树类似，思路比较清晰。

关键点在于如何寻找链表的中点，运用了快慢指针。
快指针每次走两步，慢指针每次走一步，当快指针走到最后一个节点时，慢指针刚好走了一半，也就是中间节点。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedListToBST(self, head: ListNode) -> TreeNode:
        if head is None:
            return head
        if head.next is None:
            return TreeNode(head.val)
        slow = head
        fast = head
        last = slow
        while fast.next and fast.next.next:
            last = slow
            slow = slow.next
            fast = fast.next.next
        fast = slow.next
        last.next = None
        root = TreeNode(slow.val)
        if head != slow:
            root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(fast)
        return root