【转】二元查找树转为双向链表

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。  　　比如将二元查找树                                               10                                             /      /                                           6       14                                        /   /    /　 /                                      4     8  12  16  转换成双向链表 。 4=6=8=10=12=14=16。 

声明：思想不是原创。

方法一：其思想是分别转换左右子树成双向链，再通过根节点进行连接

#include <iostream>
using namespace std;
struct BSTree
{
    int num;
    BSTree* left;
    BSTree* right;

};
BSTree *convertToLink(BSTree *tree,bool asRight)
{
    BSTree *pleft=0;//节点tree的左子树转换成的双向链表
    BSTree *pright=0;//节点tree的右子树转换成的双向链表
    //如果左子树不为空，则先转换左子树
    if(tree->left)
        pleft=convertToLink(tree->left,false);
    //如果双向左链表不为空，将此节点加到末尾
    if(pleft)
    {
        pleft->right=tree;
        tree->left=pleft;
    }
    if(tree->right)
        pright=convertToLink(tree->right,true);
    if(pright)
    {
        pright->left=tree;
        tree->right=pright;
    }
    BSTree *pTemp=tree;
    //如果当前节点tree为其根节点的右子树，则返回此树的最小节点，即为父节点的右子链表的最左节点
    if(asRight)
    {
        while(pTemp->left)
            pTemp=pTemp->left;
    }
    //否则返回最大节点，相当于父节点左子链表的最右节点
    else
    {
        while(pTemp->right)
            pTemp=pTemp->right;
    }
    return pTemp;
}
void main()
{
    BSTree *root;
    BSTree *p4 = new BSTree();
    p4->num=4;
    BSTree *p8 = new BSTree();
    p8->num=8;

    BSTree *p6 = new BSTree();
    p6->num=6;
    p6->left=p4;
    p6->right=p8;
    BSTree *p12 = new BSTree();
    p12->num=12;

    BSTree *p14 = new BSTree();
    p14->num=14;
    p14->left=p12;
    BSTree *p10 = new BSTree();
    p10->num=10;
    p10->left=p6;
    p10->right=p14;
    root = p10;
    root=convertToLink(root,true);
    cout<<root->num<<"=";
    while(root->right){
        root=root->right;
        cout<<root->num<<"=";
    }
    cin.get();
}

方法二：更容易理解，中序遍历树，那依次访问的节点本身就是有序的，将节点依次加入到链表末尾即可

#include <iostream>
using namespace std;
struct BSTree
{
    int num;
    BSTree* left;
    BSTree* right;

};
/**
中序遍历二元查找树，将节点依次加入到双向链表末尾
**/
void convertToLink2(BSTree *tree,BSTree *&lastNodeInList)
{
    if(tree->left)
        convertToLink2(tree->left,lastNodeInList);

    if(lastNodeInList)
    {
        lastNodeInList->right=tree;
        tree->left=lastNodeInList;
        lastNodeInList=tree;
    }
    else
    {
        lastNodeInList=tree;
    }
    if(tree->right)
        convertToLink2(tree->right,lastNodeInList);
}
void main()
{
    BSTree *root;
    BSTree *p4 = new BSTree();
    p4->num=4;

    BSTree *p8 = new BSTree();
    p8->num=8;

    BSTree *p6 = new BSTree();
    p6->num=6;
    p6->left=p4;
    p6->right=p8;
    BSTree *p12 = new BSTree();
    p12->num=12;


    BSTree *p14 = new BSTree();
    p14->num=14;
    p14->left=p12;

    BSTree *p10 = new BSTree();
    p10->num=10;
    p10->left=p6;
    p10->right=p14;
    root = p10;
    //root=convertToLink(root,true);
    //cout<<root->num<<"=";
    //while(root->right){
    //  root=root->right;
    //  cout<<root->num<<"=";
    //}
    BSTree *doubleLink=NULL;
    convertToLink2(root,doubleLink);
    while(doubleLink->left)
        doubleLink=doubleLink->left;
    cout<<doubleLink->num;
    while(doubleLink->right)
    {
        doubleLink=doubleLink->right;
        cout<<"="<<doubleLink->num;
    }
    cin.get();
}