程序员面试题精选100题(01)－把二元查找树转变成排序的双向链表[数据结构

转自：http://zhedahht.blog.163.com/blog/static/254111742007127104759245/

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。最后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;}