题目：输入一棵二元查找树，将该二元查找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，只调整指针的指向。

比如将二元查找树

                                        10
                                          /    \
                                        6       14
                                      /  \     /　 \
                                   　4     8  12 　  16
转换成双向链表

4=6=8=10=12=14=16。

分析：本题是微软的面试题。很多与树相关的题目都是用递归的思路来解决，本题也不例外。下面我们用两种不同的递归思路来分析。

　　思路一：当我们到达某一结点准备调整以该结点为根结点的子树时，先调整其左子树将左子树转换成一个排好序的左子链表，再调整其右子树转换右子链表。最近链接左子链表的最右结点（左子树的最大结点）、当前结点和右子链表的最左结点（右子树的最小结点）。从树的根结点开始递归调整所有结点。

　　思路二：我们可以中序遍历整棵树。按照这个方式遍历树，比较小的结点先访问。如果我们每访问一个结点，假设之前访问过的结点已经调整成一个排序双向链表，我们再把调整当前结点的指针将其链接到链表的末尾。当所有结点都访问过之后，整棵树也就转换成一个排序双向链表了。

参考代码：

首先我们定义二元查找树结点的数据结构如下：

1 struct BSTreeNode // a node in the binary search tree
2     {
3         int          m_nValue; // value of node
4         BSTreeNode  *m_pLeft;  // left child of node
5         BSTreeNode  *m_pRight; // right child of node
6     };

思路一对应的代码：

 1 ///////////////////////////////////////////////////////////////////////
 2 // Covert a sub binary-search-tree into a sorted double-linked list
 3 // Input: pNode - the head of the sub tree
 4 //        asRight - whether pNode is the right child of its parent
 5 // Output: if asRight is true, return the least node in the sub-tree
 6 //         else return the greatest node in the sub-tree
 7 ///////////////////////////////////////////////////////////////////////
 8 BSTreeNode* ConvertNode(BSTreeNode* pNode, bool asRight)
 9 {
10       if(!pNode)
11             return NULL;
12 
13       BSTreeNode *pLeft = NULL;
14       BSTreeNode *pRight = NULL;
15 
16       // Convert the left sub-tree
17       if(pNode->m_pLeft)
18             pLeft = ConvertNode(pNode->m_pLeft, false);
19 
20       // Connect the greatest node in the left sub-tree to the current node
21       if(pLeft)
22       {
23             pLeft->m_pRight = pNode;
24             pNode->m_pLeft = pLeft;
25       }
26 
27       // Convert the right sub-tree
28       if(pNode->m_pRight)
29             pRight = ConvertNode(pNode->m_pRight, true);
30 
31       // Connect the least node in the right sub-tree to the current node
32       if(pRight)
33       {
34             pNode->m_pRight = pRight;
35             pRight->m_pLeft = pNode;
36       }
37 
38       BSTreeNode *pTemp = pNode;
39 
40       // If the current node is the right child of its parent, 
41       // return the least node in the tree whose root is the current node
42       if(asRight)
43       {
44             while(pTemp->m_pLeft)
45                   pTemp = pTemp->m_pLeft;
46       }
47       // If the current node is the left child of its parent, 
48       // return the greatest node in the tree whose root is the current node
49       else
50       {
51             while(pTemp->m_pRight)
52                   pTemp = pTemp->m_pRight;
53       }
54  
55       return pTemp;
56 }
57 
58 ///////////////////////////////////////////////////////////////////////
59 // Covert a binary search tree into a sorted double-linked list
60 // Input: the head of tree
61 // Output: the head of sorted double-linked list
62 ///////////////////////////////////////////////////////////////////////
63 BSTreeNode* Convert(BSTreeNode* pHeadOfTree)
64 {
65       // As we want to return the head of the sorted double-linked list,
66       // we set the second parameter to be true
67       return ConvertNode(pHeadOfTree, true);
68 }

思路二对应的代码：

 1 ///////////////////////////////////////////////////////////////////////
 2 // Covert a sub binary-search-tree into a sorted double-linked list
 3 // Input: pNode -           the head of the sub tree
 4 //        pLastNodeInList - the tail of the double-linked list
 5 ///////////////////////////////////////////////////////////////////////
 6 void ConvertNode(BSTreeNode* pNode, BSTreeNode*& pLastNodeInList)
 7 {
 8       if(pNode == NULL)
 9             return;
10 
11       BSTreeNode *pCurrent = pNode;
12 
13       // Convert the left sub-tree
14       if (pCurrent->m_pLeft != NULL)
15             ConvertNode(pCurrent->m_pLeft, pLastNodeInList);
16 
17       // Put the current node into the double-linked list
18       pCurrent->m_pLeft = pLastNodeInList; 
19       if(pLastNodeInList != NULL)
20             pLastNodeInList->m_pRight = pCurrent;
21 
22       pLastNodeInList = pCurrent;
23 
24       // Convert the right sub-tree
25       if (pCurrent->m_pRight != NULL)
26             ConvertNode(pCurrent->m_pRight, pLastNodeInList);
27 }
28 
29 ///////////////////////////////////////////////////////////////////////
30 // Covert a binary search tree into a sorted double-linked list
31 // Input: pHeadOfTree - the head of tree
32 // Output: the head of sorted double-linked list
33 ///////////////////////////////////////////////////////////////////////
34 BSTreeNode* Convert_Solution1(BSTreeNode* pHeadOfTree)
35 {
36       BSTreeNode *pLastNodeInList = NULL;
37       ConvertNode(pHeadOfTree, pLastNodeInList);
38 
39       // Get the head of the double-linked list
40       BSTreeNode *pHeadOfList = pLastNodeInList;
41       while(pHeadOfList && pHeadOfList->m_pLeft)
42             pHeadOfList = pHeadOfList->m_pLeft;
43 
44       return pHeadOfList;
45 }

以上转自何海涛博客