Can deleting a single node in an AVL tree ever require two double rotations?

I’ve been studying AVL trees and their balancing mechanisms, specifically how rotations are used to maintain balance after insertions or deletions. I understand that a single or double rotation may be required when the balance factor of a node becomes −2 or +2.

However, I have a question regarding double rotations during a deletion operation:

Question:

Is it theoretically possible for the deletion of a single node in an AVL tree to require two double rotations to restore balance?

From my understanding, this should not be the case. Below, I’ll outline my reasoning for why only one double rotation is sufficient. If I am incorrect or missing something, I’d greatly appreciate any clarification or corrections.

My Reasoning:

Nature of AVL Tree Rotations:

When a node is deleted, the balancing operation affects only the ancestors of the deleted node along the path from the deletion point to the root. Each rotation (single or double) affects the local height of the subtree being rebalanced and restores balance for that subtree. Single Path of Propagation:

When a node is deleted, the imbalance propagates along a single path to the root. Only the first imbalanced node (let’s call it Z) encountered along this path requires rebalancing. Once Z is rebalanced, the subtree rooted at Z has its height restored, and further propagation stops.

Double Rotation Criteria:

A double rotation is triggered when:

The balance factor of Z becomes −2 or +2. The child node of Z has a balance factor in the opposite direction, requiring a two-step adjustment (e.g., left-right or right-left). After the double rotation at Z, the height of Z's subtree is restored, preventing further imbalance propagation to Z's ancestors.

Why Multiple Double Rotations Are Unnecessary:

A single double rotation restores the balance for Z and its subtree, stopping any height changes from propagating further up the tree. Thus, it should be impossible for the deletion of a single node to require more than one double rotation along the rebalancing path. Request: If my reasoning is flawed or there exist edge cases where more than one double rotation is required during a single node deletion in an AVL tree, could you please provide an explanation or counterexample?

I reviewed the AVL tree rebalancing process and tried reasoning through its propagation limits. I expected that a single deletion would affect only one imbalanced node along the path to the root, requiring at most one double rotation to restore balance. This matches my understanding of AVL tree properties, but I want to confirm if this logic holds universally or if edge cases exist.

Yes, it is possible for the deletion of a single node in an AVL tree to require two double rotations to restore balance.

In your reasoning that this is not so, you write:

A single double rotation restores the balance for Z and its subtree, stopping any height changes from propagating further up the tree.

But this is not true: the height changes are not necessarily stopped from propagating further up the tree. A double rotation at node Z reduces the height of the subtree that was first rooted by Z (which by the rotation is replaced by its grandchild node). This reduced height can surely bring an imbalance at an ancestor node, and there is no reason why that wouldn't need a double rotation: it depends on subtrees that are not part of the current subtree.

Example tree and deletion

Here is an example of a deletion that leads to two double rotations. We start with this correctly balanced AVL tree:

              __4_
             /    \
            1      9
           / \    / \
          0   3  7  10
             /  / \   \
            2  5   8  11
                \
                 6

Then we delete 0:

              __4_
             /    \
            1      9
             \    / \
              3  7  10
             /  / \   \
            2  5   8  11
                \
                 6

The node 1 now has an imbalance. As its right child is left-heavy, we need a double rotation at node 1, which results in this tree:

              __4_
             /    \
            2      9
           / \    / \
          1   3  7  10
                / \   \
               5   8  11
                \
                 6

But this double rotation has introduced an imbalance at node 4. And as its right child is left-heavy, we need a double rotation there as well:

              __4_
             /    \
            2      9
           / \    / \
          1   3  7  10
                / \   \
               5   8  11
                \
                 6

Which results in this balanced tree:

                __7_
               /    \
              4      9
             / \    / \
            2   5  8  10
           / \   \      \
          1   3   6     11

And so this proves that -- following the standard deletion procedure -- we may need to apply more than one double rotation.