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 ):
        # write code here
        if not pRootOfTree:
            return None
        arr = self.inorderTraversal(pRootOfTree)
        for i in range(len(arr)-1):
            arr[i].right = arr[i+1]
            arr[len(arr)-1-i].left = arr[len(arr)-2-i]

        return arr[0]

    def inorderTraversal(self, root):
        if not root:
            return []
        left = self.inorderTraversal(root.left)
        right = self.inorderTraversal(root.right)

        return left + [root] + right

使用一个辅助数组解决，将二叉搜索树按照中序遍历得到排序后结果，然后依次连接每个节点即可。