Walk through an AVL insertion that causes a left-left rotation.

A left-left (LL) imbalance happens when a node becomes left-heavy because a key was inserted into the left subtree of its left child. It is fixed with a single right rotation about the unbalanced node. Walk through it concretely.

Setup: insert 30, 20, 10

Insert 30, then 20 (goes left), then 10 (goes left of 20). Now 30 has balance factor -2 (left subtree height 2, right subtree height 0) — the AVL invariant is violated, and the inserted key lies in 30's left child's left subtree. That is the LL case.

        30  (bf = -2)   <- unbalanced
     /
   20   (bf = -1)
  /
10

The fix: one right rotation about 30

In a right rotation, the left child (20) becomes the new subtree root; the old root (30) becomes 20's right child. The general rule, for pivot P with left child L:

• L moves up to take P's position.
• P becomes L's right child.
• L's old right subtree (call it T) — the only orphaned link — reattaches as P's new left child. (Here T is empty.)

This preserves BST ordering: every key in L's left subtree < L < T < P < P's right subtree, which is exactly the order the rotated shape encodes.

        20   (bf = 0)
     /  \
   10    30
 (bf=0) (bf=0)

Why it works and what it costs

The rotation is O(1) — three pointer rewires (20.right = 30, 30.left = T, fix the parent link) plus updating two stored heights. It lowers the subtree's height from 3 back to 2, which restores balance not just at 20 but, because the height shrank back to its pre-insert value, at every ancestor above it too. That is why an AVL insertion needs only this one rotation and can stop climbing.

The mirror image — inserting into the right child's right subtree — is the RR case, fixed by a single left rotation. The two "outside" cases (LL, RR) need one rotation; the two "inside" cases (LR, RL) need a double rotation because a single rotation would just move the imbalance to the other side.