Question

I am building a Heap data structure, but the part that is driving me nuts is checking for a null child value.

I am using a Vector, and I am comparing a parent against a child, but if there’s only one child the program crashes since something like

vectorObject.get(i) //i'th element doesn't exist 

will return an exception.

I can’t check for the null element with something like

if (vectorObject.get(i) == null)

since running the get() method will automatically break the program, so how do you ACTUALLY check for a non-existing element without some unreadable weird hack-around??

Answer 1

You seem very confused. I think you need to re-read whatever reference you’re using to design your heap data structure (or read the relevant section of the Wikipedia article).

If your parent is at 0 then your children are at 2*(0)+1=1 and 2*(0)+2=2. In this case, 1 < vectorObject.size() will be true but 2 < vectorObject.size() is false, meaning there is a left child but no right child.

Since the Vector (or ArrayList if you’ve switched as per my suggestion) is zero-based you need to check if i < vectorObject.size(), not i <= vectorObject.size(). If i < vectorObject.size() then i is a legal index.

Update:

There are two ways you can structure the logic here. If you have to handle the single-child case totally differently than the two-child case then this would probably work best:

int size = vectorObject.size();
if (2*i+2 < size) { /* Two-child case */ }
else if (2*i+1 < size) { /* One-child case */ }
else { /* No children case */ }

If you handle the two nodes separately, then nesting might be better:

int size = vectorObject.size();
if (2*i+1 < size) {
  // Handle left child
  if (2*i+2 < size) {
    // Handle right child
  }
  else {
    // No right child
  }
}
else { /* No children */ }