1. Dynamic Dictionaries Primary Operations: Additional operations:
get(key) => search
put(key, element) => insert
remove(key) => delete
Additional operations:
ascend()
get(index)
remove(index)
Other additional ops: nearest match and range search

2. Complexity Of Dictionary Operations get(), put() and remove()
n is number of elements in dictionary
Worst-case for hash table may be made O(log n) using balanced search trees for overflow handling. C++ STL and Java use linear open addressing.

3. Complexity Of Other Operations ascend(), get(index), remove(index)
D is number of buckets
Hash table get and remove using O(n) select would take O(D + n) time.

4. The Operation put() Put a pair whose key is 35. 20 10 6 2 8 15 40 3025 35
This slide begins a review of operations on an unbalanced binary search tree.
Put a pair whose key is 35.

5. The Operation remove()
Three cases:
Element is in a leaf.
Element is in a degree 1 node.
Element is in a degree 2 node.

6. Remove From A Leaf Remove a leaf element. key = 7 20 10 6 2 8 15 40 3025 35 7 18
Remove a leaf element. key = 7

7. Remove From A Degree 1 Node20 10 6 2 8 15 40 30 25 35 7 18
Remove from a degree 1 node. key = 40

8. Remove From A Degree 1 Node (contd.)20 10 6 2 8 15 40 30 25 35 7 18
Remove from a degree 1 node. key = 15

9. Remove From A Degree 2 Node20 10 6 2 8 15 40 30 25 35 7 18
Remove from a degree 2 node. key = 10

10. Remove From A Degree 2 Node20 10 40 6 15 30 18 25 35 2 8
Replace with largest key in left subtree (or smallest in right subtree). 7

11. Remove From A Degree 2 Node20 10 40 6 15 30 18 25 35 2 8
Replace with largest key in left subtree (or smallest in right subtree). 7

12. Remove From A Degree 2 Node20 8 40 6 15 30 18 25 35 2 8
Replace with largest key in left subtree (or smallest in right subtree). 7

13. Remove From A Degree 2 Node20 8 40 6 15 30 18 25 35 2 8 7
Largest key must be in a leaf or degree 1 node.

14. Another Remove From A Degree 2 Node20 10 6 2 8 15 40 30 25 35 7 18
Remove from a degree 2 node. key = 20

15. Remove From A Degree 2 Node20 10 40 6 15 30 18 25 35 2 8 7
Replace with largest in left subtree.

16. Remove From A Degree 2 Node20 10 40 6 15 30 18 25 35 2 8 7
Replace with largest in left subtree.

17. Remove From A Degree 2 Node18 10 40 6 15 30 18 25 35 2 8 7
Replace with largest in left subtree.

18. Remove From A Degree 2 Node18 10 40 6 15 30 25 35 2 8 7
Complexity is O(height).

19. Yet Other Operations Priority Queue Motivated Operations:
find max and/or min
remove max and/or min
initialize
meld
Insert in PQ is same as insert in dictionary

20. Max And/Or Min Follow rightmost path to max element.20 10 6 2 8 15 40 30 25
Could keep a bst plus 2 variables: min and max. Now, O(1) to find min and max elements.
Follow rightmost path to max element.
Follow leftmost path to min element.
Search and/or remove => O(h) time.

21. Initialize Sort n elements.20 10 6 2 8 15 40 30 25
Sort n elements.
Initialize search tree.
Output in inorder (O(n)).
Initialize must take O(n log n) time, because it isn’t possible to sort faster than O(n log n).

22. Meld 5 1 7 17 12 10 6 2 8 15 10 6 2 1 5 8 17 15 12 7
We can merge two sorted lists by melding the corresponding binary search trees.
Create a binary search tree for each sorted listed that is to be merged0 compares (e.g., left skewed, right skewed, balanced, …)
Meld
Create merged list by inorder traversal (0 compares)
The only place compares may occur is in the meld.

23. Meld And Merge
Worst-case number of comparisons to merge two sorted lists of size n each is 2n-1.
So, complexity of melding two binary search trees of size n each is W(n).
So, logarithmic time melding isn’t possible.

24. O(log n) Height Trees = + Full binary trees. Complete binary trees.
Exist only when n = 2k –1.
Complete binary trees.
Exist for all n.
Cannot insert/delete in O(log n) time. 8 2 1 6 15 10 10 6 2 8 15
All nodes are affected by the insert. = + 1

25. Balanced Search Trees Height balanced. Weight Balanced.
AVL (Adelson-Velsky and Landis) trees
Weight Balanced.
Degree Balanced.
2-3 trees
2-3-4 trees
red-black trees
B-trees
Red-black trees are a clever binary representation of Because of time limitations, we shall not see this correspondence in class.

26. AVL Tree binary tree for every node x, define its balance factor
balance factor of x = height of left subtree of x
– height of right subtree of x
balance factor of every node x is – 1, 0, or 1

27. Balance Factors -1 1 1 -1 1 -1
This is an AVL tree.

28. Height Of An AVL Tree
The height of an AVL tree that has n nodes is at most 1.44 log2 (n+2).
The height of every n node binary tree is at least log2 (n+1).
log2 (n+1) <= height <= 1.44 log2 (n+2)

29. Proof Of Upper Bound On Height
Let Nh = min # of nodes in an AVL tree whose height is h.
N0 = 0.
N1 = 1.

30. Nh, h > 1 L R Both L and R are AVL trees. The height of one is h-1.
The height of the other is h-2.
The subtree whose height is h-1 has Nh-1 nodes.
The subtree whose height is h-2 has Nh-2 nodes.
So, Nh = Nh-1 + Nh
First, observe that N increases with h as each of Nh’s subtrees is AVL and of smaller height (at least one has height h-1).

31. Fibonacci Numbers F0 = 0, F1 = 1. Fi = Fi-1 + Fi-2 , i > 1.
N0 = 0, N1 = 1.
Nh = Nh-1 + Nh-2 + 1, i > 1.
Nh = Fh+2 – 1.
Fi ~ fi/sqrt(5).
f = (1 + sqrt(5))/2
height <= 1.44 log2 (n+2) follows
Note that for Fibonacci heaps Ni = Fi+2. Also, for Fibonacci heaps, i is the degree
(not height).

32. AVL Search Tree 1 -1 10 7 8 3 5 30 40 20 25 35 45 60
AVL search tree is often simply referred to as an AVL tree.