There are four cases and any node insertion will be one of these four cases,
1. An insertion into left subtree of the left child of a.
2. An insertion into right subtree of the left child of a.
3. An insertion into left subtree of the right child of a.
4. An insertion into right subtree of the right child of a.
Answer:
Step 1: Inserting 20
Step 2: Inserting 21
Step 3: Inserting 22
Node 20 has balance -2 so this tree is not AVL tree.
Case 4 An insertion into right sub tree of the right child of a.
Step 4: Inserting 17
Step 5: Inserting 18
Node 20 has become unbalanced
Case 2 An insertion into right sub tree of the left child of a.
An insertion into right sub tree of the left child of a.
So applying two rotations first left rotation and then right rotation.
Step 6: Inserting 23
Step 7: Inserting 27
Case 4 An insertion into right sub tree of the right child of a.
So applying single left rotation,
Step 8: Inserting 25
Still an AVL tree,
Step 9: Inserting 30
Still an AVL tree,
Step 10: Inserting 33
It is Case 4 mean insertion on right side of right node. So applying left rotation on the edge of the node where imbalance occured,
Step 11: Inserting 34
Case 4 insertion in right node of right sub tree, so applying single left rotation,
This is our Final Tree.
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