程序员面试题精选(01)－把二元查找树转变成排序的双向链表

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;
}