边工作边刷题：70天一遍leetcode: day 17-1

Convert Sorted List to Binary Search Tree

只能顺序访问链表结点，所以不能像array那样preorder的方式建树。而是inorder，因为inorder的顺序和有序的。

• 每次访问完左子树，这时候的head指向的是当前的root。这时再移动链表指针。
• 因为python是pass-by-value，而recursion的返回值为子树的root，所以只能用global variable表示当前指针（如果c++可以用pass-by-reference update链表指针）

# 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 sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        self.h = head
        n = 0
        while head:
            n+=1
            head=head.next
        return self.sorteddfs(0, n-1)
        
    def sorteddfs(self, low, high):
        if low>high: return None
        m = low+(high-low)/2
        
        left = self.sorteddfs(low, m-1)
        root = TreeNode(self.h.val)
        if self.h: self.h = self.h.next
        right = self.sorteddfs(m+1, high)
        root.left = left
        root.right = right
        return root