程式員面試題精選（1）：把二元尋找樹轉變成排序的雙向鏈表

　　題目：輸入一棵二元尋找樹，將該二元尋找樹轉換成一個排序的雙向鏈表。要求不能建立任何新的結點，只調整指標的指向。　　比如將二元尋找樹
                                            10
                                          /    /
                                        6       14
                                      /  /     /　 /
                                   　4     8  12 　  16
轉換成雙向鏈表4=6=8=10=12=14=16。　　分析：本題是微軟的面試題。很多與樹相關的題目都是用遞迴的思路來解決，本題也不例外。下面我們用兩種不同的遞迴思路來分析。　　思路一：當我們到達某一結點準備調整以該結點為根結點的子樹時，先調整其左子樹將左子樹轉換成一個排好序的左子鏈表，再調整其右子樹轉換右子鏈表。最近連結左子鏈表的最右結點（左子樹的最大結點）、當前結點和右子鏈表的最左結點（右子樹的最小結點）。從樹的根結點開始遞迴調整所有結點。　　思路二：我們可以中序遍曆整棵樹。按照這個方式遍曆樹，比較小的結點先訪問。如果我們每訪問一個結點，假設之前訪問過的結點已經調整成一個排序雙向鏈表，我們再把調整當前結點的指標將其連結到鏈表的末尾。當所有結點都訪問過之後，整棵樹也就轉換成一個排序雙向鏈表了。參考代碼：首先我們定義二元尋找樹結點的資料結構如下：
    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
    };思路一對應的代碼：
///////////////////////////////////////////////////////////////////////
// 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);
}思路二對應的代碼：
///////////////////////////////////////////////////////////////////////
// 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;
}