微软面试100题—自做

由于发现题集具有很好的借鉴效果，整个题目乃至答案如下：

http://blog.csdn.net/v_JULY_v/article/details/6057286

1.把二元查找树转变成排序的双向链表（树）
 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
};

我个人的代码： 

 struct BSTreeNode
{
  int m_nValue; // value of node
  BSTreeNode *m_pLeft; // left child of node
  BSTreeNode *m_pRight; // right child of node
};
 BSTreeNode* bstree= new BSTreeNode;//Notice this; claim it obviously
 void build_tree(BSTreeNode* bstree);//build a tree
 void convert_tree2chain(BSTreeNode* bstree);


int _tmain(int argc, _TCHAR* argv[])
{
build_tree(bstree);
return 0;
}
void build_tree(BSTreeNode* bstree)
{
int tmp=0;
int count=0;
while(1)
{


cin>>tmp;
if(tmp==27)
{break;}


if(count==0)
{
bstree->m_nValue=tmp;
}
else if(bstree->m_nValue<tmp&&bstree->m_pLeft==NULL)

//本人程序与答案者的一大区别在于条件的判断上，NULL==bstree->m_nValue，能够防止产生相应的笔误赋值错误,还是这样防错性比较好
{//insert left leaf
BSTreeNode* bs_temp=new BSTreeNode;
bs_temp->m_nValue=tmp;
bstree->m_pLeft=bs_temp;
}else if(bstree->m_nValue>tmp&&bstree->m_pRight==NULL)
{//insert right leaf
BSTreeNode* bs_temp=new BSTreeNode;
bs_temp->m_nValue=tmp;
bstree->m_pRight=bs_temp;
}else if(bstree->m_nValue<tmp&&bstree->m_pLeft!=NULL)
{//if the node is null, then go to the next node
bstree=bstree->m_pLeft;
build_tree(bstree);//作递归调用
}
else if(bstree->m_nValue>tmp&&bstree->m_pRight!=NULL)
{
bstree=bstree->m_pRight;
build_tree(bstree);//作递归调用
}




}
}
void convert_tree2chain(BSTreeNode* bstree)
{
BSTreeNode* temp=new BSTreeNode;
temp->m_pLeft=bstree->m_pLeft;
temp->m_pRight=bstree->m_pRight;
//还是建立了一个中间节点，可以查找闲置的节点达到
while(bstree->m_pLeft!=NULL)
{//find the left leaf' right leaf 实际就是找节点的左边节点的最右与右边节点的最右
if(bstree->m_pLeft->m_pRight!=NULL)
bstree->m_pLeft=bstree->m_pLeft->m_pRight;
else
{
break;
}
}
while(bstree->m_pRight!=NULL)
{//find the right leaf' left leaf
if(bstree->m_pRight->m_pLeft!=NULL)
bstree->m_pRight=bstree->m_pRight->m_pLeft;
else
{
break;
}
}
convert_tree2chain(temp->m_pLeft);
convert_tree2chain(temp->m_pRight);
}

与其中答案的对比：

#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;//节点处赋值的应该将其置为空，防止出现空指针
        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;
}