[编程之美-11]把二元查找树变成为排序的双向链表

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
};

主要考察:二叉树的中序遍历算法的思想.

代码如下:

#include<iostream>
#include<stdio.h>
using namespace std;

struct BSTreeNode
{
int m_nValue;
BSTreeNode *m_pleft;
BSTreeNode *m_pright;
};

typedef struct BSTreeNode DoubleList;
DoubleList *pHead, *pListIndex;

void convertToDoubleList(BSTreeNode *pCurrent);

void addBSTreeNode(BSTreeNode *&pCurrent, int value);

void ergodicBSTree(BSTreeNode *pCurrent);

int main()
{
BSTreeNode *pRoot = NULL;
pListIndex = NULL;
pHead = NULL;

addBSTreeNode(pRoot, 10);
addBSTreeNode(pRoot, 4);
addBSTreeNode(pRoot, 6);
addBSTreeNode(pRoot, 8);
addBSTreeNode(pRoot, 12);
addBSTreeNode(pRoot, 14);
addBSTreeNode(pRoot, 15);
addBSTreeNode(pRoot, 16);

ergodicBSTree(pRoot);

return 0;
}

void addBSTreeNode(BSTreeNode *&pCurrent, int value)
{
if(pCurrent == NULL)
{
BSTreeNode *pBSTree = new BSTreeNode();
pBSTree->m_nValue = value;
pBSTree->m_pleft = NULL;
pBSTree->m_pright = NULL;
pCurrent = pBSTree;
}
else
{
if((pCurrent->m_nValue) > value)
addBSTreeNode(pCurrent->m_pleft, value);
else if((pCurrent->m_nValue) < value)
addBSTreeNode(pCurrent->m_pright, value);
}
}

void ergodicBSTree(BSTreeNode *pCurrent)
{
if(pCurrent == NULL)
return ;

if(pCurrent->m_pleft != NULL)
ergodicBSTree(pCurrent->m_pleft);

convertToDoubleList(pCurrent);

if(pCurrent->m_pright != NULL)
ergodicBSTree(pCurrent->m_pright);
}

void convertToDoubleList(BSTreeNode *pCurrent)
{
pCurrent->m_pleft = pListIndex;

if(pListIndex != NULL)
pListIndex->m_pright = pCurrent;
else
pHead = pCurrent;

pListIndex = pCurrent;
cout<<pCurrent->m_nValue<<endl;
}