What is the worst-case performance of the insert method in an AVL tree with linked nodes?

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Let’s evaluate what the insert operation does in an AVL tree with linked nodes and how its time scales with n.
Option 1: O(1) — This would imply a constant-time operation regardless of the tree size. In an AVL tree, insertion potentially requires traversing from the root to a leaf to find the insertion point, and possibly performing rebalancing rotations. Therefore, it cannot be constant time as n grows.
Option 2: O(n) — This would mean the insertion time grows linearly with the number of nodes. While a simple unbalanced BST could degrade this badly, AVL trees maintain balance, so the worst-case height does not grow linearly with n; it stays logarithmic, making O(n) too large for AVL insert.
Option 3: O(log n) — This reflects the height-bounded nature of AVL trees. The insertion path goes from root to a leaf along a path of length equal to the tree height, which is O(log n) due to AVL balancing. Each rotation or pointer adjustment is constant time, so the total remains proportional to the height.
Option 4: O(n^2) — This would indicate a quadratic-time operation, which would be true only in scenarios with repeated expensive operations or highly degenerate structures. AVL trees prevent that kind of growth in insert cost, so this is incorrect.
In summary, the correctness of each option hinges on how the tree’s height scales with n and how rotations contribute to the cost. The balancing property constrains height to O(log n), and the work done at each level (pointer changes and a constant number of rotations) does not explode beyond that bound.

CMPSC 132 Spring 2025 Module 8 Checkpoint

What is the worst-case performance of the insert method in an AVL tree with linked nodes?

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