[coolpad]把二元查找树转变成排序的双向链表

原文链接：http://www.cnblogs.com/wolenski/archive/2012/07/08/2581859.html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 nodeBSTreeNode  *m_pLeft;  // left child of nodeBSTreeNode  *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-treeif(pNode->m_pLeft)pLeft = ConvertNode(pNode->m_pLeft, false);// Connect the greatest node in the left sub-tree to the current nodeif(pLeft){pLeft->m_pRight = pNode;pNode->m_pLeft = pLeft;}// Convert the right sub-treeif(pNode->m_pRight)pRight = ConvertNode(pNode->m_pRight, true);// Connect the least node in the right sub-tree to the current nodeif(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 nodeif(asRight){while(pTemp->m_pLeft4000)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 nodeelse{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 truereturn 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-treeif (pCurrent->m_pLeft != NULL)ConvertNode(pCurrent->m_pLeft, pLastNodeInList);// Put the current node into the double-linked listpCurrent->m_pLeft = pLastNodeInList;if(pLastNodeInList != NULL)pLastNodeInList->m_pRight = pCurrent;pLastNodeInList = pCurrent;// Convert the right sub-treeif (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 listBSTreeNode *pHeadOfList = pLastNodeInList;while(pHeadOfList && pHeadOfList->m_pLeft)pHeadOfList = pHeadOfList->m_pLeft;return pHeadOfList;}

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