使用递归函数打印链表

Question

我可以打印我的链接列表：

|
|-> [ID: 3] Poetry
|   |
|   |-> [ID: 3] Notes 3
|   |   
|   |-> [ID: 2] Notes 2
|   |   
|   |-> [ID: 1] Notes 1
|   |   
|
|-> [ID: 2] Diet
|   |
|   |-> [ID: 2] Diet 2
|   |   
|   |-> [ID: 1] Diet 1
|   |   
|
|-> [ID: 1] Program

使用我编写的以下函数：

void printn(NODE *handle)   //only 2 depths allowed
    {
    NODE *copy_handle = NULL; 
    NODE *depth1 = NULL;
    NODE *depth2 = NULL;

    for( copy_handle = handle ; copy_handle != NULL ; copy_handle = copy_handle->pNext  )
        {
        printf("|\n|-> [ID: %d] %s\n", copy_handle->id, copy_handle->pLabel);

        if(copy_handle->pInner != NULL )
            {
            printf("|\t|\n");
            for(depth1 = copy_handle->pInner ; depth1 != NULL ; depth1 = depth1->pNext  )
                {
                printf("|\t|-> [ID: %d] %s\n", depth1->id, depth1->pLabel);

                if(depth1->pInner != NULL )
                    {
                    printf("|\t|\t|\n");
                    for(depth2 = depth1->pInner ; depth2 != NULL ; depth2 = depth2->pNext  )
                        {
                        printf( "|\t|\t|-> [ID: %d] %s\n", depth2->id, depth2->pLabel);
                        }
                    }

                printf("|\t|\t\n");
                } 
            }
        }

    }

然而，此函数的局限性在于我只能打印节点的子节点和该子节点的子节点，或者正如我在代码中所称的那样，我只能允许 depth = 2。我想做的是能够打印无限深度，所以我查看了我的原始函数，我觉得用递归重新设计它是合适的。所以我开发了以下内容：

void rec_printn(NODE *handle, int tab)   //unlimited depths
    {

    printf("tab %d\n", tab); //for testing purposes
    NODE *copy_handle = handle;
    for( ; copy_handle != NULL ; copy_handle = copy_handle->pNext )
        {
        printf("%*s|-> [ID: %d] %s\n", tab, " ",  copy_handle->id, copy_handle->pLabel);   //add variable spacing 

        if(copy_handle->pInner != NULL )
            {
            tab+=5;
            rec_printn(copy_handle->pInner , tab);
            }
        else if(copy_handle->pNext == NULL )
            {
            tab-=5;
            printf("<take back the indent\n"); //for testing purposes
            }

        }
    }

我在这里遇到的问题是缩进没有像我在上面的原始示例中所期望的那样返回，而是得到以下内容：

tab 5
     |-> [ID: 3] Poetry
tab 10
          |-> [ID: 3] Notes 3
          |-> [ID: 2] Notes 2
          |-> [ID: 1] Notes 1
<take back the indent
          |-> [ID: 2] Diet
tab 15
               |-> [ID: 2] Diet 2
               |-> [ID: 1] Diet 1
<take back the indent
               |-> [ID: 1] Program
<take back the indent

问题：我做错了什么，缩进不是他们应该的那样？

Answer 1

您在每次迭代中将 5 添加到 tab。

不是改变 tab，而是将正确的值传递给函数并让递归处理它：

if (copy_handle->pInner != NULL)
{
    rec_printn(copy_handle->pInner, tab + 5);
}
else if (copy_handle->pNext == NULL )
{
    printf("<take back the indent\n");
}

这将确保 tab 始终具有与 "this level" 相同的值。

（也就是说，不要想 "increase the indentation, then print the next level"，想 "print the next level with a deeper indentation"，如果你明白我的意思。）

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