算法面试100题——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 node
BSTreeNode *m_pLeft; // left child of node
BSTreeNode *m_pRight; // right child of node
};

/*用途：**说明：**算法：*///#define LOCAL#include <cstdio>#include <cstdlib>#include <stack>using namespace std;struct BSTreeNode{  int m_nValue;  BSTreeNode *m_pLeft;  BSTreeNode *m_pRight;};typedef BSTreeNode* BST;BSTreeNode* NewNode();BSTreeNode* NewNode(int v);//递归实现BSTreeNode* ToDLinkList(BST bst);void Traverse(BST bst, BSTreeNode* &pre);//显示栈实现 BSTreeNode* ToDLinkList2(BST bst); void TestHead(BSTreeNode* head);void TestTail(BSTreeNode* tail);int main(){#ifdef LOCAL    freopen(".in", "r", stdin);    freopen(".out", "w", stdout);#endif    BST bst = NewNode();    BSTreeNode *n0 = NewNode(10);    bst->m_pLeft = n0;    BSTreeNode *n1 = NewNode(6);    n0->m_pLeft = n1;    BSTreeNode *n2 = NewNode(14);    n0->m_pRight = n2;    BSTreeNode *n3 = NewNode(4);    n1->m_pLeft = n3;    BSTreeNode *n4 = NewNode(8);    n1->m_pRight = n4;    BSTreeNode *n5 = NewNode(12);    n2->m_pLeft = n5;    BSTreeNode *n6 = NewNode(16);    n2->m_pRight = n6;    BSTreeNode* tail = bst->m_pLeft;    //测试尾指针     while(tail->m_pRight)        tail = tail->m_pRight;    BSTreeNode* head;//  head = ToDLinkList(bst);    //递归实现    head = ToDLinkList2(bst);   //显示栈实现    TestHead(head);    printf("\n");    TestTail(tail);    return 0;}BSTreeNode* NewNode(){    BSTreeNode* pNode = (BSTreeNode*)malloc(sizeof(BSTreeNode));    pNode->m_pLeft = pNode->m_pRight = NULL;    return pNode;}BSTreeNode* NewNode(int v){    BSTreeNode* pNode = (BSTreeNode*)malloc(sizeof(BSTreeNode));    pNode->m_nValue = v;    pNode->m_pLeft = pNode->m_pRight = NULL;    return pNode;}//递归实现BSTreeNode* ToDLinkList(BST bst){    BSTreeNode* head = bst;    while(head->m_pLeft)        head = head->m_pLeft;    BSTreeNode* pre = NULL;     //当前节点的前一个指针     Traverse(bst->m_pLeft, pre);    return head;}void Traverse(BST bst, BSTreeNode* &pre){    if(bst != NULL){        Traverse(bst->m_pLeft, pre);        bst->m_pLeft = pre;        if(pre)            pre->m_pRight = bst;        pre = bst;        Traverse(bst->m_pRight, pre);    }}//显示栈实现BSTreeNode* ToDLinkList2(BST bst){    stack<BSTreeNode*> s;    BSTreeNode* head = bst;    while(head->m_pLeft)        head = head->m_pLeft;    BSTreeNode* pre = NULL;    BSTreeNode* p = bst->m_pLeft;    while(!s.empty() || p){        if(p){            s.push(p);            p = p->m_pLeft;        }        else{            BSTreeNode* m_pTop = s.top();            s.pop();            m_pTop->m_pLeft = pre;            if(pre)                pre->m_pRight = m_pTop;            pre = m_pTop;            if(m_pTop->m_pRight)                p = m_pTop->m_pRight;        }    }    return head;} void TestHead(BSTreeNode* head){    while(head){        printf("%d ", head->m_nValue);        head = head->m_pRight;    }}void TestTail(BSTreeNode* tail){    while(tail){        printf("%d ", tail->m_nValue);        tail = tail->m_pLeft;    }}