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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 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: the head of the sub tree // Output: the tail of the double-linked list /////////////////////////////////////////////////////////////////////// BSTreeNode* ConvertNode(BSTreeNode* pNode) { if(pNode == NULL) return NULL;
BSTreeNode *pLastNodeInList = NULL; BSTreeNode *pCurrent = pNode;
// Convert the left sub-tree if (pCurrent->m_pLeft != NULL) pLastNodeInList = ConvertNode(pCurrent->m_pLeft);
// 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) pLastNodeInList = ConvertNode(pCurrent->m_pRight);
return pLastNodeInList; }
/////////////////////////////////////////////////////////////////////// // Covert a binary search tree into a sorted double-linked list // Input: the head of // Output: the head of sorted double-linked list /////////////////////////////////////////////////////////////////////// BSTreeNode* Convert(BSTreeNode* pHeadOfTree) { BSTreeNode *pLastNodeInList = ConvertNode(pHeadOfTree);
// 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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来自： ShangShujie > 《SourceCode》
