1. 二元查找树转变成排序的双向链表
1.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。
????? 10
???? /??? \
??? 6??? 14
? ?/ \??? / \
? 4 8 12 16
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转换成双向链表
4=6=8=10=12=14=16
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首先我们定义的二元查找树 节点的数据结构如下：
struct BSTreeNode{ int m_nValue; // value of node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node};?
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题目分析：?
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。
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思路一对应的代码：
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;} 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：
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。
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;}?
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