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。

举例：

10
/   \        转变
6     14      -->   4=6=8=10=12=14=16
/ \   /  \
4   8 12   16

定义的二元查找树结点的数据结构如下：

struct BSTreeNode
{
int m_nValue;
BSTreeNode *m_pLeft;
BSTreeNode *m_pRight;
};

答：用二叉树的中序遍历

#include "stdafx.h"
#include <iostream>
#include <fstream>

using namespace std;

struct BSTreeNode
{
int m_nValue;
BSTreeNode *m_pLeft;
BSTreeNode *m_pRight;
};

//创建二元二叉树
void CreateBitree(BSTreeNode *&pNode, fstream &fin)
{
int dat;
fin>>dat;
if(dat==0)
{
pNode = NULL;
}
else
{
pNode = new BSTreeNode();
pNode->m_nValue=dat;
CreateBitree(pNode->m_pLeft, fin);
CreateBitree(pNode->m_pRight, fin);
}
}

//中序递归转变
void change(BSTreeNode *pNode, BSTreeNode *&pTail)
{
if (NULL != pNode)
{
change(pNode->m_pLeft, pTail);
if (NULL != pTail)
{
pTail->m_pRight = pNode;
}
pNode->m_pLeft = pTail;
pTail = pNode;
change(pNode->m_pRight, pTail);
}
}

int _tmain(int argc, _TCHAR* argv[])
{
BSTreeNode *pRoot = NULL;
BSTreeNode *pTail = NULL;
fstream fin("tree.txt");
CreateBitree(pRoot, fin);
change(pRoot, pTail);
while(NULL != pTail)
{
cout<<pTail->m_nValue<<"  ";
pTail = pTail->m_pLeft;
}

cout<<endl;
return 0;
}

运行的界面如下：



建造二叉树用到的tree.txt为：

10 6 4 0 0 8 0 0 14 12 0 0 16 0 0