把二元查找树转变成排序的双向链表(微软面试题)

参考了July的博客，然后自己动手实现了一线，留着备忘吧！

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。

实现如下：

#include <iostream>
#include <cstdio>
using namespace std;

struct BSTreeNode  //存储二叉查找树节点信息
{
    int val;
    BSTreeNode *lchild,*rchild;
}; 

typedef BSTreeNode DoubleList; 

DoubleList *pHead,*pLastIndex; //pHead存储头节点，pLaseIndex存储上一个插入的节点

//中序序列创建二叉查找树
void CreateBSTree(BSTreeNode * & pCurrent,int val)  //pCurrent是一个BSTreeNode*类型的引用
{
    if(NULL == pCurrent)
    {
        BSTreeNode *temp=new BSTreeNode() ;
        temp->val = val ;
        temp->lchild = NULL , temp->rchild = NULL ;
        pCurrent = temp ;  
    }
    else
    {
        if(pCurrent->val > val)
            CreateBSTree(pCurrent->lchild,val);
        else if(pCurrent->val < val)
            CreateBSTree(pCurrent->rchild,val);
        else cout<<"输入数据重复"<<endl;
    }
}

void ConvertToDoubleList(BSTreeNode *node);  //将二叉树转换为双向链表

//中序遍历二叉查找树
void InorderBSTree(BSTreeNode *pCurrent)
{
    if(pCurrent == NULL) return;
    if(pCurrent->lchild != NULL)
        InorderBSTree(pCurrent->lchild);
    ConvertToDoubleList(pCurrent);
    if(pCurrent->rchild != NULL)
        InorderBSTree(pCurrent->rchild);
}

void ConvertToDoubleList(BSTreeNode *node)
{
    node->lchild = pLastIndex ;
    if(pLastIndex == NULL)
        pHead = node ;
    else
        pLastIndex->rchild = node ;
    pLastIndex = node ;  
    cout<<" "<<node->val<<endl;  
}

int main()
{
    BSTreeNode *pRoot ;
    pRoot = pHead = pLastIndex = NULL ;
    CreateBSTree(pRoot,4);
    CreateBSTree(pRoot,10); 
    CreateBSTree(pRoot,6);
    CreateBSTree(pRoot,16);
    CreateBSTree(pRoot,14);
    CreateBSTree(pRoot,12);
    CreateBSTree(pRoot,8);
    InorderBSTree(pRoot); 
    return 0;
}