Insert 10, 11, 7, 8, 13, 6 and 12 into an empty AVL tree. The resulting tree is:

Question

BlackTom AI · Accepted Answer

The task asks to insert the keys 10, 11, 7, 8, 13, 6, and 12 into an empty AVL tree and determine the resulting structure.

First, insert 10. The tree is simply:
10

Next, insert 11. It goes to the right of 10. The tree is now balanced with 10 as root and 11 as its right child:
  10
    \
    11

Then insert 7. It goes to the left of 10. The tree remains balanced since the left and right subtrees of 10 have height 1 each:
    10
   /  \
  7   11

Now insert 8. Since 8 > 7, it becomes the right child of 7. The subtree rooted at 7 has a left height of 0 and right height of 1, so 7 is still balanced, and 10’s two subtrees have heights 2 (via 7-8) and 1 (via 11), so 10 is balanced overall:
    10
   /  \
  7   11
   \
    8

Insert 13. It goes to the right of 11. The right subtree of 10 becomes a bit taller, but the balance factor at every node remains within [-1, 1], so the tree is still balanced:
    10
   /  \
  7   11
   \     \
    8     13

Insert 6. It goes to the left of 7. The subtree rooted at 7 now has both left and right children (6 and 8). The height of that subtree becomes 2, while the right subtree of 10 (rooted at 11) has height 2 (11 -> 13). The overall tree stays balanced:
    10
   /  \
  7   11
 / \     \
6   8     13

Finally, insert 12. It goes to the right of 11, then to the left of 13 (since 12 < 13). This insertion makes 11’s right subtree heavier and creates an RL (Right-Left) imbalance at node 11 (and hence at 10 as well). To restore balance, perform a right rotation on the subtree rooted at 13, followed by a left rotation on the subtree rooted at 11.

- Right-rotate at 13: the subtree with root 13 becomes 12 as the right child of 11, with 13 becoming the right child of 12.
- Left-rotate at 11: the node 12 becomes the right child of 10, with 11 as the left child of 12 and 13 as the right child of 12.

After these rotations, the final AVL structure is:
        10
       /  \
      7    12
     / \   / \
    6  8  11 13

This matches the AVL property (each node’s left and right subtrees differ in height by at most 1) and reflects the correct rebalancing following the RL case on inserting 12. Note that there were no provided multiple-choice options to compare against, but the derived tree is the correct final AVL after all insertions.

CMPSC 132 Spring 2025 Module 6.1 Checkpoint

Insert 10, 11, 7, 8, 13, 6 and 12 into an empty AVL tree. The resulting tree is:

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