题解 | #二叉搜索树与双向链表#

python递归指针，非列表

#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

#
# 
# @param pRootOfTree TreeNode类 
# @return TreeNode类
#
class Solution:
    def __init__(self):
        self.head = TreeNode(0)
        self.cur = self.head
    def Convert(self , pRootOfTree ):
        if not pRootOfTree:
            return
        self.midorder(pRootOfTree)
        cur1 = self.head.right
        self.head.right.left = None
        return cur1

    def midorder(self, node):
        if node == None:
            return
        self.midorder(node.left)
        self.cur.right = TreeNode(node.val)
        self.cur.right.left = self.cur
        self.cur = self.cur.right
        self.midorder(node.right)