1. AVL TREE Name :TIN HO

2. Introduction An AVL tree is another balanced binary search tree. An AVL tree is another balanced binary search tree. Named after their inventors, Adelson-Velskii and Landis, Named after their inventors, Adelson-Velskii and Landis, They were the first dynamically balanced trees to be proposed. They were the first dynamically balanced trees to be proposed. Like red-black trees, they are not perfectly balanced, but pairs of sub-trees differ in height by at most 1, maintaining an O(logn) search time. Like red-black trees, they are not perfectly balanced, but pairs of sub-trees differ in height by at most 1, maintaining an O(logn) search time.

3. Definition of an AVL tree ► Addition and deletion operations also take O(logn) time. ► An AVL tree is a binary search tree which has the following properties: ► The sub-trees of every node differ in height by at most one. ► Every sub-tree is an AVL tree.

4. Balance requirement for an AVL tree: the left and right sub-trees differ by at most 1 in height 1.T h e s u b - t r e e s o f e v e r y n o d e d i f f e r i n h e i g h t b y a t m o s t o n e. 2.E v e r y s u b - t r e e i s a n A V L t r e e.

5. Be careful with this definition: it permits some apparently unbalanced trees! For example, here are some trees: Tree AVL tree? Yes Examination shows that each left sub-tree has a height 1 greater than each right sub-tree.

6. ► AVL tree? ► No ► Sub-tree with root 8 has height 4 and sub-tree with root 18 has height 2

7. Key terms AVL trees Trees which remain balanced - and thus guarantee O(logn) search times - in a dynamic environment. Or more importantly, since any tree can be re-balanced - but at considerable cost - can be re-balanced in O(logn) time. Trees which remain balanced - and thus guarantee O(logn) search times - in a dynamic environment. Or more importantly, since any tree can be re-balanced - but at considerable cost - can be re-balanced in O(logn) time.

8. AVL TREE AVL Tree An AVL tree is a tree which is balanced We earlier defined a balanced tree as a tree whose height is [log (n+1)] Another definition is “A tree is balanced if the number of nodes in every left sub-tree differs by at most 1 from the number of nodes in the corresponding right sub-tree. This is known as count-balanced One method of count-balancing a tree is to repeatedly move the root into the sub-tree with the smaller amount of nodes Each move is called a shift

9. Gary Eric John Jill Carl Mary Fred Ann Hank Kim Derek Unbalanced Tree

10. This tree has 7 nodes to the left of the root and only 3 nodes to the right of the root To count-balance we will right shift Jill into the right sub-tree Jill’s in-order predecessor, Hank will now become the new root

11. Unbalanced Tree After 1 Shift Right Jill Eric John Hank Carl Mary Fred Ann Gary Kim Derek

12. Unbalanced Tree After 1 Shift Right The tree is still unbalanced with 6 nodes to the left of root and 4 nodes to the left The tree is still unbalanced with 6 nodes to the left of root and 4 nodes to the left We perform a shift right one more time moving Hank into the right tree and making his in-order predecessor (Gary) the new root We perform a shift right one more time moving Hank into the right tree and making his in-order predecessor (Gary) the new root

13. Unbalanced Tree After 2 Shifts Right n The tree is now balanced 5-5. However the Root->left subtree is not balanced i.e 3-1 so to correct that we will right shift Eric Jill Eric John Gary Carl Mary HankAnn Fred Kim Derek

14. 1 Shift Right On Root Left SubTree Jill Eric John Gary Carl Mary HankAnn Fred Kim Derek n Now we have an AVL Tree

15. Count Balance Algorithm // return tree whose root is n, as a count balanced (AVL) tree public BinNode countbalance(BinNode n) { if (not n empty) then leftnum = number of nodes in left child rightnum = number of nodes in right child if (leftnum > rightnum) then loop for i going from1 to (leftnum-rightnum)/2 do right-shift root else loop for i going from1 to (rightnum-leftnum)/2 do left-shift root endif n left child = countbalance(n left child) n right child = countbalance(n right child) return n }