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 <stdio.h>
#include <iostream.h>

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"<<endl;
        }
    }
}

// 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<<pCurrent->m_nValue<<endl;
}

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;
}
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Press any key to continue
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struct BSTreeNode{int m_nValue;//value of nodeBSTreeNode *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_nValue<value)addBSTreeNode(pCurrent->m_pRight,value);elsecout<<"加入了重复节点"<<endl;}}void ergodicBSTree(BSTreeNode* pCurrent){if(NULL==pCurrent)return;if(NULL!=pCurrent->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;}elsepHead=pCurrent;pListIndex=pCurrent;cout<<pCurrent->m_nValue<<endl;}int main(void){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);while(pHead){cout<<pHead->m_nValue<<" ";pHead=pHead->m_pRight;}cout<<endl;while(pListIndex){cout<<pListIndex->m_nValue<<" ";pListIndex=pListIndex->m_pLeft;}return 0;}