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
};*/

#include "stdafx.h"
#include <iostream>
using namespace std;
struct BSTreeNode
{
	int m_nValue; // value of node
	BSTreeNode *m_pLeft; // left child of node
	BSTreeNode *m_pRight; // right child of node
};

typedef BSTreeNode DoubleList;
DoubleList *pHead;
DoubleList *pListIndex;

void convertToDoubList(BSTreeNode *pCurrent);

//创建二叉树
void addBSTreeNode(BSTreeNode *&pCurrent,int value)
{
	
                  //如果树节点为NULL,创建新的节点
                  if (pCurrent == NULL)
	{
		BSTreeNode *BSNode = new BSTreeNode();
		BSNode->m_pRight = NULL;
		BSNode->m_pLeft = NULL;
		BSNode->m_nValue = value;
		pCurrent = BSNode;
	}else
	{	
		if(pCurrent->m_nValue > value)
		{
			addBSTreeNode(pCurrent->m_pLeft,value);
		}else 
		{
			addBSTreeNode(pCurrent->m_pRight,value);
		}
	}
}
//中序遍历，相当于一个排序算法，先遍历左子树，在遍历根节点，最后遍历右子树
void ergodicBSTree(BSTreeNode *pCurrent)
{
	if (pCurrent != NULL)
	{
		ergodicBSTree(pCurrent->m_pLeft);
		 convertToDoubList(pCurrent);
		//cout << pCurrent->m_nValue;
		//cout << " ";
		ergodicBSTree(pCurrent->m_pRight);
	
	}
}
void convertToDoubList(BSTreeNode *pCurrent)
{
	pCurrent->m_pLeft = pListIndex;
	if (NULL != pListIndex)
	{
		pListIndex->m_pRight = pCurrent;//双向的~
	}else
	{
		pHead = pCurrent;
	}
	pListIndex = pCurrent;
	cout << pCurrent->m_nValue << " ";
}

int _tmain(int argc, _TCHAR* argv[])
{
	BSTreeNode *mRoot = NULL;
	addBSTreeNode(mRoot,10);
	addBSTreeNode(mRoot,8);
	addBSTreeNode(mRoot,6);
	addBSTreeNode(mRoot,4);
	addBSTreeNode(mRoot,12);
	addBSTreeNode(mRoot,14);
	addBSTreeNode(mRoot,16);
	ergodicBSTree(mRoot);
	system("pause");
	return 0;
}