Java Recursive Call (Novice)

public void preorder(int index) {
        // Create Content
        if(aTree[index] != null) {
        System.out.print(aTree[index]);
        preorder(2 * index);
        preorder(2 * index + 1);
        }
    }

Simple tree potential traversal function. Array is a 32-size object array with an input index of 1. Null value from aTree[12]. What I'm curious about is that after debugging, when the index becomes 16 and the if statement becomes false, Preorder (2 * index + 1); it goes to this function. What's the reason? Shouldn't the preorder function end automatically because the if statement is false?
Why

preorder(2 * index); 
(index == 8) -> 
preorder(16);
if(aTree[index] != null)  == false -> 
preorder(2 * index + 1);  -> 
preorder(17) -> 
if(aTree[index] != null)  == false -> 
preorder(2 * index + 1); -> 
preorder(4)

What's the reason for the function to go over? I don't understand it in my head

java recursive

2022-09-20 16:32

1 Answers

In a given function, the preorder function is called twice in total, once in 2 * index and once in 2 * index+1. So this is how the program works, right?

preorder(1) ->preorder(2) ->preorder(4) ->preorder(8) ->preorder(16) ->null ->preorder(17)->null ->preorder(9) ->... ->... ->preorder(3)-> ...

So when if is false, the function ends and then you go to process the remaining functions that you still have to run. I don't know if it's explained because I'm not good at speaking.

2022-09-20 16:32

If you have any answers or tips

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