题目描述
输入一棵二叉搜索树，现在要将该二叉搜索树转换成一个排序的双向链表。而且在转换的过程中，不能创建任何新的结点，只能调整树中的结点指针的指向来实现。
二叉搜索树，左子结点的值总是小于父结点的值，右子结点的值总是大于父结点的值。
解题思路
如果用中序遍历二叉搜索树，遍历结点的顺序就是从小到大。所以这题的基本思路就是：用中序遍历一下二叉树，同时保留当前链表的最后一个指针pLast，以及当前指针pCurrent，每访问一个节点更新一下两者的值。
代码实现stdlibtime.h>using namespace std;typedef struct TreeNode{int data;TreeNode *Leftchild;TreeNode *Rightchild;}TreeNode, *PtreeNode;PtreeNode InsertSortTree(PtreeNode Tree, int data){if(Tree == NULL){Tree = new TreeNode;Tree;Tree->Leftchild = NULL;Tree->Rightchild = NULL;})Tree->Leftchild = InsertSortTree(Tree->Leftchild, data);)Tree->Rightchild = InsertSortTree(Tree->Rightchild, data);return Tree;}PtreeNode CreatSortTree(PtreeNode Tree){srand(time(NULL));cout endl;for(int i ; i ++){int ;cout ;Tree = InsertSortTree(Tree, data);}cout << endl;return Tree;}void PreOrderTraverse(PtreeNode Tree){if(Tree == NULL)return;PreOrderTraverse(Tree->Leftchild);cout ;PreOrderTraverse(Tree->Rightchild);}void CoverToList(PtreeNode Tree, PtreeNode *lastNode){if(Tree == NULL)return;PtreeNode pCurrent = Tree;if(pCurrent->Leftchild != NULL)CoverToList(pCurrent->Leftchild, lastNode);pCurrent->Leftchild = *lastNode;if((*lastNode) != NULL)(*lastNode)->Rightchild = pCurrent;*lastNode = pCurrent;if(pCurrent->Rightchild != NULL)CoverToList(pCurrent->Rightchild, lastNode);}PtreeNode Cover(PtreeNode Tree){PtreeNode lastNode = NULL;CoverToList(Tree, &lastNode);PtreeNode pHeadList = lastNode;pHeadList->Leftchild != NULL)pHeadList = pHeadList->Leftchild;return pHeadList;}void Print_Tree(PtreeNode Tree){cout endl;PreOrderTraverse(Tree);}void Print_List(PtreeNode List){cout endl;){cout ;Rightchild;}cout << endl;}int main(int argc, char const *argv[]){PtreeNode Tree = NULL;Tree = CreatSortTree(Tree);Print_Tree(Tree);Tree = Cover(Tree);Print_List(Tree);return 0;}
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