输出如下图形

*

**
***
****
*****
****
***
**

*

这道题本来也不难。

CSDN的牛们也各有奇招。

但正好理解OSG的node遍历，因此就写了一个旁门左道的。 

#include <iostream>



class node

{

public:

    node(void){;}

    ~node(void){;}

    virtual void print(void) = 0;

};



class group:public node

{

public:

    group(void):_child(NULL){;}

    ~group(void){ delete _child;}

    void print(void)

    {

        curdepth++;

        for (int j=0; j<curdepth; j++)

        {

            printf("*");

        }

        if (_child)

        {

            printf("\n");

            _child->print();

        }

        else

        {

            printf("\n");

        }

        if (_child)

        {

            for (int j=0; j<curdepth; j++)

            {

                printf("*");

            }

            printf("\n");

        }

        curdepth--;

    }

    node * _child;

    static int curdepth;

};

int group::curdepth = 0;





int main(int argc, char** argv)

{

    int asteriskmaxnum = 5;

    group * ptr = NULL;



    group * iter = ptr;

    for (int i =0; i<asteriskmaxnum; i++)

    {

        if (!iter)

        {

            ptr = new group;

            iter = ptr;

        }

        else

        {

            iter->_child = new group;

            iter = (group *)(iter->_child);

        }

    }



    ptr->print();

    system("PAUSE");

    return 0;

}