按照左右子树分治,递归实现。根的左边连接左子树的最右边结点,右边连接右子树的最左边结点。

'''
输入一棵二叉搜索树,将该二叉搜索树转换成一个排序的双向链表。
要求不能创建任何新的结点,只能调整树中结点指针的指向。
'''

# -*- coding:utf-8 -*-
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    def Convert(self, pRootOfTree):
        if pRootOfTree == None:
            return None
        if not pRootOfTree.left and not pRootOfTree.right:
            return pRootOfTree

        # 处理左子树
        self.Convert(pRootOfTree.left)
        left = pRootOfTree.left

        # 连接根与左子树最大结点
        if left:
            while left.right:
                left = left.right
            pRootOfTree.left, left.right = left, pRootOfTree

        # 处理右子树
        self.Convert(pRootOfTree.right)
        right = pRootOfTree.right

        # 连接根与右子树最小结点
        if right:
            while right.left:
                right = right.left
            pRootOfTree.right, right.left = right, pRootOfTree

        while pRootOfTree.left:
            pRootOfTree = pRootOfTree.left

        return pRootOfTree

pNode1 = TreeNode(8)
pNode2 = TreeNode(6)
pNode3 = TreeNode(10)
pNode4 = TreeNode(5)
pNode5 = TreeNode(7)
pNode6 = TreeNode(9)
pNode7 = TreeNode(11)

pNode1.left = pNode2
pNode1.right = pNode3
pNode2.left = pNode4
pNode2.right = pNode5
pNode3.left = pNode6
pNode3.right = pNode7

S = Solution()
newList = S.Convert(pNode1)
print(newList.val)