18.3. Handling Nodes with Height +2

So, the third situation is that it is not 0 or plus 1, but the slope at whiles minus 1. So, we have already dealt with the case where 0 or plus 1 is the possibility. So, if it is minus 1 this means that the right sub tree must be strictly taller than the next left sub tree.

In an AVL tree, maintaining balance is crucial for efficient operations. When we identify that a node has a height difference (or balance factor) of +2, it indicates that the left subtree is significantly taller than the right subtree. This specific scenario pertains to when the balance factor is -1, meaning that while the right subtree is shorter, we still need to acknowledge that the left subtree is notably taller. Thus, any adjustments will involve rotations to restore balance.

Think of a seesaw where one side (the left subtree) is considerably heavier (taller) than the other (the right subtree). If too much weight is placed on one side, it will tip over, causing an imbalance. To correct this, we need to shift some weight (perform a tree rotation) to maintain equilibrium.

Now, this h plus 2 it now beings split of y and the rest therefore h plus 1 must be coming in the right and h 1 the left we got the slope by assumption we are assuming slope is minus 1.

When dealing with the height of the nodes, if we have a node with height +2, we can say that the left child has a height of h+1. Therefore, the subtrees on the right must balance this out. The assumption of a slope or balance factor of -1 shows how the right subtree is shorter but needs to be adjusted for balance. By ensuring we keep track of these layers of height differences, we can effectively optimize the tree structure.

Consider a tall stack of books. If one side has two extra books (making it 'h + 2'), while the other side only has one (h + 1), the stack will lean. To make it level, we either add books to the shorter side or redistribute them, which relates to the balancing acts in the AVL tree when nodes are adjusted.

So, now I have a two-step procedure what I will do is, now I will do similar rotation, but to the left involving these two nodes. So, I will hang up these sub trees here by z.

The process of balancing an unbalanced AVL tree primarily involves rotations. When we decide to rotate left, it means that we will take the current node and adjust it downwards while promoting the left child upwards. This effectively redistributes the heights of the subtrees, ensuring that nodes remain balanced after the rotation. We temporarily treat the left child as a new root for those rotations.

Imagine stacking blocks. If one block is too tall and leaning over, you can take it from the top and place it on the side, creating a new base. This is much like how rotations work in balancing trees, where we are shifting positions to restore a perfectly balanced stack.

Now, if you go back and check all the heights, then you find that at y I either have height 0 or I have height plus 1. Now, if I look at z then the left hand side is now h plus 1 and the right hand side is h or h minus 1.

After performing rotations, it's critical to check the heights of the nodes to ensure that they maintain their balancing properties. The goal is to have each node's balance factor at -1, 0, or +1. By checking the heights, we determine if we have successfully restored the balance and if further adjustments are required.

Think about keeping a till in a store balanced. If you keep adding and removing change, you must verify the weights are even on both sides. Just like adjusting balances on either side of the checkout, we do the same checks in our AVL tree—ensuring it's balanced each time we add or perform operations.

So, to summarize this what we have said is that if we have plus 2, then we look at the left child of x call it y, if the slope of y is 0 or 1 we rotate at right at x.

To summarize the process of addressing nodes that present with a height of +2, we initially inspect the left child of the node. Depending on the balance factor of this subtree (whether it’s 0 or 1), we can determine a necessary single right rotation. If the balance factor is -1, we perform a left rotation first at y, followed by right rotation at x to ensure a balanced tree structure.

Consider a large seesaw where kids on different sides must be balanced. If one group is heavier but a bit unstable, we either shift a kid from the heavier side to balance the seesaw or move a few kids on the lighter side down to stabilize it. This is similar to the rotations we use in the AVL tree management.

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