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Insert 30, 20, 25 into an AVL tree. Which rotations happen, and why?

Walk through AVL balance factors, the four imbalance cases, and single and double rotations.

TRT practice prompt — not a verified question from a named employer.

The Runtime Theory Team2 min read
Learn it firstAVL rotationsPro — members onlyAn AVL tree keeps the heights of every node's two subtrees within one of each other. After an insert, at most one or two rotations, local pointer rearrangements that preserve key order, restore that rule and keep the height below 1.44 log₂ n.

A strong answer

A strong candidate frames Insert 30, 20, 25 into an AVL tree. Which rotations happen, and why? around the exact memory layout, compiler semantics, and hardware execution model covered in AVL rotations. Walk through AVL balance factors, the four imbalance cases, and single and double rotations.

Georgy Adelson-Velsky and Evgenii Landis published the first self-balancing binary search tree in 1962, and the AVL tree is named after them. Its rule is simple: at every node, the heights of the left and right subtrees differ by at most one. The difference, height(left) − height(right), is the node's balance factor, and it must be −1, 0 or +1.

Each node stores its own height, recomputed from its children's heights whenever they change, so balance factors are available in O(1). After an insert or delete, only the nodes on the path from the change back to the root can have changed height, so only they can become unbalanced, and the algorithm checks them on the way back up.

In a production interview, always connect the specification rule to its concrete runtime consequence: state what lives on the stack versus the heap or static storage segment, how the compiler or operating system validates alignments, lifetimes, and error codes, and what happens under adversarial or boundary inputs (such as zero-length buffers, integer overflow, concurrent mutation, or partial I/O). Backing your explanation with a small trace of register/memory state transitions immediately separates senior systems engineers from candidates who only memorized syntax.

What the interviewer is listening for

  • Mastery of core invariant: State the AVL balance invariant and compute balance factors.
  • Mastery of core invariant: Perform single and double rotations, and choose between them from the shape of the imbalance.
  • Mastery of core invariant: Explain why the invariant bounds the height at about 1.44 log₂ n.
  • Hardware and ABI awareness: Explaining how sizes, alignments, calling conventions, or kernel boundaries behave on real 64-bit targets rather than relying on implementation-defined assumptions.
  • Defensive engineering: Naming the exact compiler flags (-Wall -Wextra -Werror), static analyzers, or runtime sanitizers (-fsanitize=address,undefined,thread) that verify correctness in CI.

Common mistakes

  • Explaining only surface syntax without tracing the underlying memory addresses, ownership lifetime, or kernel/user-space state transitions.
  • Assuming undefined behavior (such as signed overflow, uninitialized reads, strict-aliasing violations, or dangling pointers) will "just work" because it passed an unoptimized -O0 debug build.
  • Ignoring return codes, partial reads/writes, alignment padding, or cleanup paths when an allocation or syscall fails midway through a function.

This answer walks

Practice follow-ups

  1. 01Why is at most one rotation step needed after an insert?
  2. 02How many rotations can a delete need?
  3. 03Prove the 1.44 log₂ n height bound.
  4. 04How would you add subtree sizes to support k-th smallest queries?

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