题解 | #二叉搜索树与双向链表#

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

#
# 
# @param pRootOfTree TreeNode类 
# @return TreeNode类
#
class Solution:
    head = None
    pre = None 
    def Convert(self , pRootOfTree ):
        if pRootOfTree is None:
            return None
        self.Convert(pRootOfTree.left)
        if not self.pre:
            self.head = pRootOfTree
            self.pre = pRootOfTree
        else:
            self.pre.right = pRootOfTree
            pRootOfTree.left = self.pre
            self.pre = pRootOfTree
        self.Convert(pRootOfTree.right)
        return self.head