把二元查找树转变成排序的双向链表[数据结构]

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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。

03	// Input: pNode - the head of the sub tree

04	//        asRight - whether pNode is the right child of its parent

05	// Output: if asRight is true, return the least node in the sub-tree

06	//         else return the greatest node in the sub-tree

07	///////////////////////////////////////////////////////////////////////

08	BSTreeNode* ConvertNode(BSTreeNode* pNode, bool asRight)

09

{

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

}

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：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。

03	// Input: pNode -           the head of the sub tree

04	//        pLastNodeInList - the tail of the double-linked list

05	///////////////////////////////////////////////////////////////////////

06	void ConvertNode(BSTreeNode* pNode, BSTreeNode*& pLastNodeInList)

07

{

08	if(pNode == NULL)

09

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

}

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