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:
    def Convert(self , pRootOfTree ):
        res=[]
        if pRootOfTree is None:
                return None
        def zhongxu(root):
            if root is None:
                return
            zhongxu(root.left)
            res.append(root)
            zhongxu(root.right)
        zhongxu(pRootOfTree)
        for i in range(len(res)-1):
            res[i+1].left=res[i]
            res[i].right=res[i+1]
        return res[0]
##### 这题的思路就是先中序遍历一下，可以到中序节点的数组，在循环一下得到双向链表！