Construction of Minimal AVL Trees

An AVL (Adelson-Velsky and Landis) tree is a binary search tree where the difference in heights of left and right subtrees of any node is at most 1. The number of nodes in a minimal AVL tree of height $h$ is given by the recurrence relation:

$N(h) = N(h-1) + N(h-2) + 1$ where $N(h)$ represents the number of nodes in a minimal AVL tree of height $h$, and the base cases are:

$N(0) = 1 \quad \text{(a single node tree)}$ $N(1) = 2 \quad \text{(root with one child)}$

Now let's construct minimal AVL trees of heights 0, 1, 2, 3, 4, and 5 using this recurrence relation.

Minimal AVL Trees:

1. Height 0: The minimal AVL tree has 1 node, $N(0) = 1$.
2. Height 1: The minimal AVL tree has 2 nodes, $N(1) = 2$.
3. Height 2: Using the recurrence relation: $N(2) = N(1) + N(0) + 1 = 2 + 1 + 1 = 4$
4. Height 3: $N(3) = N(2) + N(1) + 1 = 4 + 2 + 1 = 7$
5. Height 4: $N(4) = N(3) + N(2) + 1 = 7 + 4 + 1 = 12$
6. Height 5: $N(5) = N(4) + N(3) + 1 = 12 + 7 + 1 = 20$

Number of nodes in a minimal AVL tree of height 6:

Using the same recurrence relation:

$N(6) = N(5) + N(4) + 1 = 20 + 12 + 1 = 33$

Thus, the number of nodes in a minimal AVL tree of height 6 is 33.

Would you like further clarification or details on this explanation?

Follow-up questions:

1. What is the relationship between height and balance in an AVL tree?
2. How does the balance condition affect the efficiency of an AVL tree?
3. Can we derive an upper bound for the height of an AVL tree with $n$ nodes?
4. What are the advantages of using AVL trees compared to unbalanced binary search trees?
5. How do rotations help maintain the balance in an AVL tree?

Tip:

In an AVL tree, rebalancing occurs after insertion or deletion, using single or double rotations to ensure that the tree remains balanced.