在Java中跟踪递归方法时遇到问题

Question

This is a rather quick question. I have been having trouble understanding the concept of recursion in Java and I was wondering if someone could help me. I have been tracing successfully 5 out of the 6 recursive methods that I have done, but this one is throwing me for a loop, literally. I was expecting the output to be just 2(the output I traced), but when I put it into my compiler it came out with 1213121(proper output). As I said, I have been able to do this thus far, but this one is confusing me. Here is what I am working with:

public class Recursion
{
    public static void main(String [] args)
    {
        Recursion r = new Recursion();
        r.doSomething(3);
    }

    public void doSomething(int n)
    {
        if (n > 0)
        {
            doSomething(n-1);
            System.out.print(n);
            doSomething(n-1);
        }
    }
}

Answer 1

I would prefer to do this in the comments, but the formatting of the comments make it hard to follow. You can follow this recursion as if it were a math equation.

The basic logic is:

DoSomething(n) results in:

• doSomething(n-1)
• print n
• doSomething(n-1)

So let's follow this for doSomething(3)

• doSomething(2)
• print 3
• soSomething(2)

So now we have to figure out what doSomething(2) does, so just plug in the values with n=2:

• doSomething(1)
• print 2
• doSomething(1)

Now plug in the values for n=1:

• doSomething(0)
• print 1
• doSomething(0)

doSomething(0) is the base case, where the recursion stops. Basically doSomething(0) does nothing.

Therefore, the actions for n=1 become

• print 1

Therefore, the actions for n=2 become

• print 1
• print 2
• print 1

Therefore, the actions for n=3 become

• print 1
• print 2
• print 1
• print 3
• print 1
• print 2
• print 1

Answer 2

Here is the diagramatic representation of how the control flows through this recursion program.

I think the diagram should clear your doubts.

Answer 3

Let's give line numbers to below code to ease further explanation.

1. doSomething(n-1);
2. System.out.print(n);

3. doSomething(n-1);

   line 1 redirect call until you reach at n=1. In the end line will call doSomething(0). because n>0, it will return and will print 1. Then again line 3 will call doSomething(0) and will return to previous caller which was n=2. It will print 2 and will call doSomething(1). Which will print 1 (121) and will return to n=2,line3. From here it will return to previous caller where n=3. Line 2 will print 3(1213). Then line 3 will call doSomething(2) which will append 121 as previous call which will ultimately become 1213121.

I hope this helps.

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