 So in this case, I've gotten a little bit more of a complex tree going on here. And what if, say for example, I wanted to give it the removal operation of Remove 3? Well, since we're in a binary tree, I would first need to find that 3, so I would traverse downward until I found it. And given that's an in, the first step would be I remove in. All right, excellent. Done, haha. Well, you see that now I have an empty spot, and I happen to have two children that I need to be working with. So what I have to decide is which node would go up. And in our case, what we can do is we can work off of what we would classify as our in-order predecessor, predecessor. The idea behind this is if we remember in-order, I look at my left children, then myself, then my right children. So in our case, we happen to have removed self from the equation. But what's the next available step? Next in the sense that I happen to have a left child that I can work off of. I happen to have a two. So what I'm going to do is I'm going to move that in-order predecessor, hopefully I can make this in the right amount of space. I move that in-order predecessor to the removed in position. And so as we can see, I would move my in-order predecessor. That's my two in-order predecessor. You'll see that word a few more times. I'm joking. I would move that to my three. So now I would sort of remove that two. And instead of my three, if I come in and I happen to erase this a little bit, let me increase the size, just a hair. There we go. I would, once again, I'd come in and I would put that two here. So now I've taken that two. I've moved it to its removed in position. And then I splay that predecessor, splay predecessor. So in our case, once again, as we've seen a few times now, I have an x, I have a y, a y, and I have a z. And in this case, I happen to see that it's in a zig, zig approach. So as we've seen in the past, I will change this so that it's an x, y, and z. But you notice that I happen to have a few children going on here. So if we sort of watched our previous videos, that t1 will go to x, that t2 will go to y, and z will get t3 and t4. So in our case, we see that t1, well, guess why? We just happen to move that. So it's gone. However, we do have a t2. We don't have a t3, but we do have a t4. So in our case, if we follow this rule of removing n, we did. We moved the in-order predecessor to that removed location. We did. We splay the predecessor now. So in our case, we see I would change it so that the root node in this case is 2. And then I would move my y to be a right child of my x. So 25 would be here. My z would then be the right child of my y. Now I also have to deal with my t1, t2, t3, and t4. So as you can see, y is going to get t2. So that means that y, 25 in this case, will get my 10. And my z will get my t4. So my z will get my t4.