java - 有人可以解释这个程序的控制流程吗？

Question

这是我的 AP 计算机科学类(class)中的示例程序，我无法理解它的控制流程。

public static void mystery( int n )
{
   System.out.println( "mystery called with n = " + n );

   if ( n == 0 )
   {
      System.out.println( "n is zero so no more recursive calls!" );
      return;
   }

   mystery( n - 1 );

   System.out.println( "We did it again with n = " + n );
}

public static void main( String[] args ) 
{
   mystery( 5 );
}

它输出这个:

mystery called with n = 5
mystery called with n = 4
mystery called with n = 3
mystery called with n = 2
mystery called with n = 1
mystery called with n = 0
n is zero so no more recursive calls!
We did it again with n = 1
We did it again with n = 2
We did it again with n = 3
We did it again with n = 4
We did it again with n = 5

到目前为止，我了解了递归方法以及它如何通过以下方式记忆起自己:

mystery( n - 1 );

但是，我看不出它如何在以下之后输出这五个语句:

n is zero so no more recursive calls!

从逻辑上讲，它似乎只会声明:

We did it again with n = 0

任何人都可以帮助学生并向我解释它是如何输出它做了什么的吗？

Answer 1

当函数完成时，调用它的函数有机会完成并执行更多代码。

这是递归如何发生的说明。递归的每个级别都由缩进的增加表示。

mystery(5):
    "mystery called with n = 5"
    mystery(4):
        "mystery called with n = 4"
        mystery(3):
            "mystery called with n = 3"
            mystery(2):
                "mystery called with n = 2"
                mystery(1):
                    "mystery called with n = 1"
                    mystery(0):
                        "mystery called with n = 0"
                        "n is zero so no more recursive calls!"
                        mystery(0) returns
                    "We did it again with n = 1"
                    end of mystery(1)
                "We did it again with n = 2"
                end of mystery(2)
            "We did it again with n = 3"
            end of mystery(3)
        "We did it again with n = 4"
        end of mystery(4)
    "We did it again with n = 5"
    end of mystery(5)