原文地址：http://zhedahht.blog.163.com/blog/static/254111742007127104759245/
題目：輸入一棵二元查找樹，將該二元查找樹轉(zhuǎn)換成一個(gè)排序的雙向鏈表。要求不能創(chuàng)建任何新的結(jié)點(diǎn)，只調(diào)整指針的指向。
比如將二元查找樹
10
/ \
6 14
/ \ / \
4 8 12 16
轉(zhuǎn)換成雙向鏈表
4=6=8=10=12=14=16。
分析：本題是微軟的面試題。很多與樹相關(guān)的題目都是用遞歸的思路來解決，本題也不例外。下面我們用兩種不同的遞歸思路來分析。
思路一：當(dāng)我們到達(dá)某一結(jié)點(diǎn)準(zhǔn)備調(diào)整以該結(jié)點(diǎn)為根結(jié)點(diǎn)的子樹時(shí)，先調(diào)整其左子樹將左子樹轉(zhuǎn)換成一個(gè)排好序的左子鏈表，再調(diào)整其右子樹轉(zhuǎn)換右子鏈表。最近鏈接左子鏈表的最右結(jié)點(diǎn)（左子樹的最大結(jié)點(diǎn)）、當(dāng)前結(jié)點(diǎn)和右子鏈表的最左結(jié)點(diǎn)（右子樹的最小結(jié)點(diǎn)）。從樹的根結(jié)點(diǎn)開始遞歸調(diào)整所有結(jié)點(diǎn)。
思路二：我們可以中序遍歷整棵樹。按照這個(gè)方式遍歷樹，比較小的結(jié)點(diǎn)先訪問。如果我們每訪問一個(gè)結(jié)點(diǎn)，假設(shè)之前訪問過的結(jié)點(diǎn)已經(jīng)調(diào)整成一個(gè)排序雙向鏈表，我們?cè)侔颜{(diào)整當(dāng)前結(jié)點(diǎn)的指針將其鏈接到鏈表的末尾。當(dāng)所有結(jié)點(diǎn)都訪問過之后，整棵樹也就轉(zhuǎn)換成一個(gè)排序雙向鏈表了。
參考代碼：
首先我們定義二元查找樹結(jié)點(diǎn)的數(shù)據(jù)結(jié)構(gòu)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
};
思路一對(duì)應(yīng)的代碼：
///////////////////////////////////////////////////////////////////////
// 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);
}
思路二對(duì)應(yīng)的代碼：
///////////////////////////////////////////////////////////////////////
// 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;
}