把二元查找树转变成排序的双向链表

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/*--------------------------------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转换成双向链表4=6=8=10=12=14=16。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 nodeBSTreeNode *m_pLeft; // left child of nodeBSTreeNode *m_pRight; // right child of node};*/#include<iostream>#include<cstdio>using namespace std;typedef struct BSTreeNode{int m_nValue; // value of nodestruct BSTreeNode *m_pLeft; // left child of nodestruct BSTreeNode *m_pRight; // right child of node}BSTreeNode;typedef BSTreeNode DoubleList;DoubleList *pHead=NULL;DoubleList *pTail=NULL;void addBSTNode(BSTreeNode *&pCurrent,int value);void convertToDoubleList(BSTreeNode *pCurrent);void inorderTraverse(BSTreeNode *&pCurrent);void addBSTNode(BSTreeNode *&pCurrent,int value){if(pCurrent==NULL){BSTreeNode *pNew=new BSTreeNode();pNew->m_nValue=value;pNew->m_pLeft=NULL;pNew->m_pRight=NULL;     pCurrent=pNew;  /*这条语句非常重要,当创建的一个新的节点，就需要把它连接到合适的位置（即成为其父节点的左孩子或右孩子）此时pCurrent作为右值是指向它的孩子节点的指针（是左孩子还是右孩子取决于调用时的实参），而作为左值就代表其存放孩子节点位置的存储区域*/}else{if(value > pCurrent->m_nValue)addBSTNode(pCurrent->m_pRight,value);else if(value < pCurrent->m_nValue)addBSTNode(pCurrent->m_pLeft,value);elsecout<<"输入的数字产生重复"<<endl;}}void convertToDoubleList(BSTreeNode *pCurrent){pCurrent->m_pLeft=pTail;if(pTail==NULL){pHead=pCurrent;pCurrent->m_pRight=NULL;}else{pTail->m_pRight=pCurrent;}pTail=pCurrent;cout<<pCurrent->m_nValue<<endl;}void inorderTraverse(BSTreeNode *&pCurrent){if(pCurrent==NULL)return;if(pCurrent->m_pLeft!=NULL)inorderTraverse(pCurrent->m_pLeft);convertToDoubleList(pCurrent);if(pCurrent->m_pRight!=NULL)inorderTraverse(pCurrent->m_pRight);}int main(){BSTreeNode *pRoot=NULL;addBSTNode(pRoot, 10);addBSTNode(pRoot, 4);addBSTNode(pRoot, 6);addBSTNode(pRoot, 8);addBSTNode(pRoot, 12);addBSTNode(pRoot, 14);addBSTNode(pRoot, 15);addBSTNode(pRoot, 16);inorderTraverse(pRoot);return 0;}
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