题解 | #二叉搜索树与双向链表#

该题看起来就与二叉树的中序遍历有相似之处，所以先写一个中序遍历，改一改就是答案了，如下所示

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

#
# 
# @param pRootOfTree TreeNode类 
# @return TreeNode类
#
class Solution:
    def Convert(self , pRootOfTree ):
        if not pRootOfTree:
            return pRootOfTree
        stack = []
        node = pRootOfTree
        result=None
        pre=None
        now=None
        
        while node or len(stack)>0:
            while node:
                stack.append(node)
                node = node.left
            if len(stack)>0:
                now =stack.pop()
                if not pre:
                    result =now
                if pre:
                    pre.right=now
                    now.left=pre
                pre = now
                node = now.right
        return result