Pointers in AVL Tree Rotation

Question

I have trouble understanding why the below tree rotation code works. If T2 points to y.left and y.left points to x, doesn't this make the last assignment x.right = T2 equal to x.right = x? Shouldn't the pointer point to the initial T2?

Node leftRotate(Node x) {
    Node y = x.right;
    Node T2 = y.left;

    // Perform rotation
    y.left = x;
    x.right = T2;

    //  Update heights
    x.height = max(height(x.left), height(x.right)) + 1;
    y.height = max(height(y.left), height(y.right)) + 1;

    // Return new root
    return y;
}

score 0 · Answer 1 · 2017-05-31T23:52:31.890

The best way to reason with this is to draw it out and go through it a step at a time:

At the beginning of the function we have Node x:

   x
  /  \
 L    R
     /   \
    l1    r1

Now we say y = R.

  R
 /  \
L1   R1

Node T2 = R.Left which is l2

Then the rotation:

y.left (R1.Left) = x

     R
   /   \
  x      R1
 / \
l   R
   /  \
  l1    r1

x.right = T2 (l1)

    R
   /  \
  x    r1
 / \
l   l1

A great resource I found for all things tries (albeit in C) is eternally confuzzled

score 0 · Accepted Answer · answered May 31 '17 at 23:40

Node T2 = y.left;

This line makes T2 point to the same place y.left pointed at the time that line was run. If y.left is updated to point to a different object -- x, in this case -- that change is not reflected in T2.

Mind you, if someone changed a property of that object, that change would be reflected. E.g. the code

Node T2 = y.left;
y.left.foo = bar;

then T2.foo would reflect the change to bar. It's changes to what y.left is referencing that aren't reflected. This is a pretty universal thing in Java, related to the whole "references are passed by value" thing.

Pointers in AVL Tree Rotation

2 Answers2