微软面试100题 第一题 把二元查找树转变成排序的双向链表

http://blog.csdn.net/v_JULY_v/article/details/6126406

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
};

//引用 245 楼 tree_star 的回复
#include
#include

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 convertToDoubleList(BSTreeNode * pCurrent);
// 创建二元查找树
void addBSTreeNode(BSTreeNode * & pCurrent, int value)
{
    if (NULL == pCurrent)
    {
        BSTreeNode * pBSTree = new BSTreeNode();
        pBSTree->m_pLeft = NULL;
        pBSTree->m_pRight = NULL;
        pBSTree->m_nValue = value;
        pCurrent = pBSTree;

    }
    else
    {
        if ((pCurrent->m_nValue) > value)
        {
            addBSTreeNode(pCurrent->m_pLeft, value);
        }
        else if ((pCurrent->m_nValue) < value)
        {
            addBSTreeNode(pCurrent->m_pRight, value);
        }
        else
        {
            //cout<<"重复加入节点"<         }
    }
}

// 遍历二元查找树  中序
void ergodicBSTree(BSTreeNode * pCurrent)
{
    if (NULL == pCurrent)
    {      
        return;
    }
    if (NULL != pCurrent->m_pLeft)
    {
        ergodicBSTree(pCurrent->m_pLeft);  
    }

    // 节点接到链表尾部
    convertToDoubleList(pCurrent);
    // 右子树为空
    if (NULL != pCurrent->m_pRight)
    {
        ergodicBSTree(pCurrent->m_pRight);
    }
}

// 二叉树转换成list
void  convertToDoubleList(BSTreeNode * pCurrent)
{

    pCurrent->m_pLeft = pListIndex;
    if (NULL != pListIndex)
    {
        pListIndex->m_pRight = pCurrent;
    }
    else
    {
        pHead = pCurrent;
    }  
    pListIndex = pCurrent;
    cout<m_nValue< }

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;
}
///
4
6
8
10
12
14
15
16
Press any key to continue
//

struct BSTreeNode
{
	int m_nValue;//value of node
	BSTreeNode *m_pLeft;//
	BSTreeNode *m_pRight;
};
typedef BSTreeNode DoubleList;
DoubleList* pHead;
DoubleList* pListIndex;
void convertToDoubleList(BSTreeNode* pCurrent);
void addBSTreeNode(BSTreeNode* &pCurrent,int value)
{
	if(NULL==pCurrent)
	{
		BSTreeNode *pBSTree=new BSTreeNode();
		pBSTree->m_pLeft=NULL;
		pBSTree->m_pRight=NULL;
		pBSTree->m_nValue=value;
		pCurrent=pBSTree;
	}
	else
	{
		if((pCurrent->m_nValue)>value)
		{
			addBSTreeNode(pCurrent->m_pLeft,value);
		}
		else if(pCurrent->m_nValuem_pRight,value);
		else
			cout<<"加入了重复节点"<m_pLeft)
		ergodicBSTree(pCurrent->m_pLeft);
	convertToDoubleList(pCurrent);
	if(NULL!=pCurrent->m_pRight)
		ergodicBSTree(pCurrent->m_pRight);
}
void convertToDoubleList(BSTreeNode* pCurrent)
{
	pCurrent->m_pLeft=pListIndex;
	if(NULL!=pListIndex)
	{
		pListIndex->m_pRight=pCurrent;
	}
	else
		pHead=pCurrent;
	pListIndex=pCurrent;
	cout<m_nValue<m_nValue<<" ";
		pHead=pHead->m_pRight;
	}
	cout<m_nValue<<" ";
		pListIndex=pListIndex->m_pLeft;
	}
	return 0;
}