Printing the values on each level of a Binary Tree
This is good code.

General

While it does what it says it will do, here is one beef I have with it, but it's a small one:


    Node temp = que.poll();
    if(temp != mark)
    System.out.print(temp.key);



No braces, no indenting, code meaning is confused.....

That should be:

    Node temp = que.poll();
    if(temp != mark) {
        System.out.print(temp.key);
    }


There is a fair amount of debate about using '1-liner' if statements. The following are reasons why I despise them:


the meaning of the code becomes white-space defined (like your indenting problem above). There are whole languages ( cough Pythong cough ) which are defined by whitespace. Java is defined by braces {}. Use them.
If you use a version-control-system like CVS, SVN, Git, etc. then small changes to code blocks can become visually big in a change-log. Adding one line to the code means updating 3 lines.
it can lead to bugs where people do not put braces in to lines where there should be two statements.
etc.
etc.


I don't like 1-liners Sam-I-Am

OK. But, as code goes, yours is good. How can it be better?

Improvements


You can make mark a private static final instance:

private static final Node MARK = new Node(0);

The following statement has redundancy:


        if(que.peek() == mark || que.isEmpty()){
            return;
        }



it could be:

        if(que.isEmpty()){
            return;
        }


It is not possible for the queue to have two consecutive marks, so the peek can never be a mark.
Now, back to that indentation on the println() method call... here's your code:


    if(temp != mark)
    System.out.print(temp.key);
    if(temp == mark){
        if(que.peek() == mark || que.isEmpty()){
            return;
        }
        que.add(mark);
        System.out.println();
    }



if you had correct braces, etc. you would see that you would be better off with an if/else condition, and that will save an if check:

    if(temp != mark) {
        System.out.print(temp.key);
    } else {
        if(que.peek() == mark || que.isEmpty()){
            return;
        }
        que.add(mark);
        System.out.println();
    }

System.out.println(...) and all the print*(...) variants are slow (they are synchronized). Your code would be faster if you used a system like:

StringBuilder sb = new StringBuilder();
....

    if (temp != mark) {
        sb.append(temp.key);
    } else {
        ....
        sb.append("\n");
    }

....
System.out.print(sb);



Conclusion

Otherwise, this is good code.