把二元查找树转变成排序的双向链表

把二元查找树转变成排序的双向链表

用的比较笨的办法，思路是按中序递归得到顺序然后依次修改其指针域。希望得到更好的算法，望知道的赐教

//二叉排序树变为双向链表
#include<iostream>
#include<stack>
using namespace std;
struct treenode{          //树节点定义
    int data;
    treenode *l;
    treenode *r;
};
void create(treenode *&t){  //建树
    t->data=10;
    t->l=new treenode();
    t->l->data=6;
    t->l->l=new treenode();
    t->l->l->data=4;
    t->l->l->l=NULL;
    t->l->l->r=NULL;
    t->l->r=new treenode();
    t->l->r->data=8;
    t->l->r->l=NULL;
    t->l->r->r=NULL;

    t->r=new treenode();
    t->r->data=14;
    t->r->l=new treenode();
    t->r->l->data=12;
    t->r->l->l=NULL;
    t->r->l->r=NULL;
    t->r->r=new treenode();
    t->r->r->data=16;
    t->r->r->l=NULL;
    t->r->r->r=NULL;

}
void printtree(treenode *t){  //测试
    if(t){
        printtree(t->l);
        cout<<t->data<<" ";
        printtree(t->r);
    }
}
void convert(treenode *&t){   //转换，主要思路为利用堆栈存储节点顺序然后逆序输出
    stack<treenode *> s,save;
    treenode *p=t;
    while(p||!s.empty()){
        if(p){s.push(p);p=p->l;}
        else{
            p=s.top();
            s.pop();
            save.push(p);
            p=p->r;
        }
    }
    treenode *pre,*next;
    next=save.top();
    save.pop();
    next->r=NULL;
    while(!save.empty()){
        pre=save.top();
        save.pop();
        pre->r=next;
        next->l=pre;
        next=pre;
    }
    next->l=NULL;
    t=next;
    
}
int main(void){
    treenode *t=new treenode();
    treenode *pre=NULL;
    create(t);
    printtree(t);
    cout<<endl;
    convert(t);
    cout<<endl;
    while(t){
        cout<<t->data<<" ";
        t=t->r;
    }
    system("pause");
    return 0;
}