把二元查找树转变成排序的双向链表

//c/c++

/*
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。
*/

#include "stdafx.h"
#include <iostream>
using namespace std;

struct TreeNode
{
int m_nValue; // value of node
TreeNode *m_pLeft; // left child of node
TreeNode *m_pRight; // right child of node
};

typedef TreeNode DoubleList;
DoubleList * pHead;
DoubleList * pListIndex;

//生成查找树
void adjust(TreeNode * & pCurrent, int value)
{
//根节点的处理，生成根节点,为pCurrent赋值  <span style="color:#FF0000;">重点理解：注意为传址，修改了main内的pRoot</span>
if(NULL == pCurrent)
{
TreeNode *pEnd = new TreeNode();
pEnd ->m_pLeft =NULL;
pEnd ->m_pRight = NULL;
pEnd ->m_nValue = value;
pCurrent = pEnd;
}
//
else
{   //在左子树添加
if (value<pCurrent->m_nValue)
{
adjust(pCurrent->m_pLeft,value);
}
//在右子树添加
else if(value>pCurrent->m_nValue)
{
adjust(pCurrent->m_pRight,value);
}
else
{
cout<<" 重 复 的 节 点 "<<endl;
}

}

}

void addList(TreeNode *pCurrent);
//covert the tree to link
void covertTreeToLink(TreeNode *pCurrent)
{
//中序读出
if(pCurrent==NULL)
{
return;
}

//遍历左子树
if(NULL != pCurrent->m_pLeft)
{
covertTreeToLink(pCurrent->m_pLeft);
}

//将根节点加进链表
addList(pCurrent);

//遍历右子树
if(NULL != pCurrent->m_pRight)
{
covertTreeToLink(pCurrent->m_pRight);
}
}
//添加节点到链表
void addList(TreeNode *pCurrent)
{

if(NULL ==pListIndex)
pHead = pListIndex;
else
{
pCurrent->m_pLeft = pListIndex;
pListIndex ->m_pRight = pCurrent;
pListIndex = pCurrent;//指针后移
}
cout<<"Now "<<pCurrent->m_nValue<<endl;
}

int main()
{   TreeNode * pRoot = NULL;
pListIndex = NULL;//链表的移动指针
pHead = NULL;//链表的头指针
adjust(pRoot, 10);
adjust(pRoot, 4);
adjust(pRoot, 6);
adjust(pRoot, 8);
adjust(pRoot, 12);
adjust(pRoot, 14);
adjust(pRoot, 15);
adjust(pRoot, 16);
covertTreeToLink(pRoot);
getchar();
return 0;
}