二元查找树转换成一个排序的双向链表

　　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：

struct BSTreeNode // a node in the binary search tree

{

int m_nValue; // value of node

BSTreeNode *m_pLeft; // left child of node

BSTreeNode *m_pRight; // right child of node

};

思路一对应的代码：

///////////////////////////////////////////////////////////////////////

// Covert a sub binary-search-tree into a sorted double-linked list

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

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

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

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

///////////////////////////////////////////////////////////////////////

BSTreeNode* ConvertNode(BSTreeNode* pNode, bool asRight)

{

if(!pNode)

return NULL;

BSTreeNode *pLeft = NULL;

BSTreeNode *pRight = NULL;

// Convert the left sub-tree

if(pNode->m_pLeft)

pLeft = ConvertNode(pNode->m_pLeft, false);

// Connect the greatest node in the left sub-tree to the current node

if(pLeft)

{

pLeft->m_pRight = pNode;

pNode->m_pLeft = pLeft;

}

// Convert the right sub-tree

if(pNode->m_pRight)

pRight = ConvertNode(pNode->m_pRight, true);

// Connect the least node in the right sub-tree to the current node

if(pRight)

{

pNode->m_pRight = pRight;

pRight->m_pLeft = pNode;

}

BSTreeNode *pTemp = pNode;

// If the current node is the right child of its parent,

// return the least node in the tree whose root is the current node

if(asRight)

{

while(pTemp->m_pLeft)

pTemp = pTemp->m_pLeft;

}

// If the current node is the left child of its parent,

// return the greatest node in the tree whose root is the current node

else

{

while(pTemp->m_pRight)

pTemp = pTemp->m_pRight;

}

return pTemp;

}

///////////////////////////////////////////////////////////////////////

// Covert a binary search tree into a sorted double-linked list

// Input: the head of tree

// Output: the head of sorted double-linked list

///////////////////////////////////////////////////////////////////////

BSTreeNode* Convert(BSTreeNode* pHeadOfTree)

{

// As we want to return the head of the sorted double-linked list,

// we set the second parameter to be true

return ConvertNode(pHeadOfTree, true);

}

思路二对应的代码：

///////////////////////////////////////////////////////////////////////

// Covert a sub binary-search-tree into a sorted double-linked list

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

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

///////////////////////////////////////////////////////////////////////

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

{

if(pNode == NULL)

return;

BSTreeNode *pCurrent = pNode;

// Convert the left sub-tree

if (pCurrent->m_pLeft != NULL)

ConvertNode(pCurrent->m_pLeft, pLastNodeInList);

// Put the current node into the double-linked list

pCurrent->m_pLeft = pLastNodeInList;

if(pLastNodeInList != NULL)

pLastNodeInList->m_pRight = pCurrent;

pLastNodeInList = pCurrent;

// Convert the right sub-tree

if (pCurrent->m_pRight != NULL)

ConvertNode(pCurrent->m_pRight, pLastNodeInList);

}

///////////////////////////////////////////////////////////////////////

// Covert a binary search tree into a sorted double-linked list

// Input: pHeadOfTree - the head of tree

// Output: the head of sorted double-linked list

///////////////////////////////////////////////////////////////////////

BSTreeNode* Convert_Solution1(BSTreeNode* pHeadOfTree)

{

BSTreeNode *pLastNodeInList = NULL;

ConvertNode(pHeadOfTree, pLastNodeInList);

// Get the head of the double-linked list

BSTreeNode *pHeadOfList = pLastNodeInList;

while(pHeadOfList && pHeadOfList->m_pLeft)

pHeadOfList = pHeadOfList->m_pLeft;

return pHeadOfList;

}