1. Binary Search Trees1 ADT for Map: Map stores elements (entries) so that they can be located quickly using keys. Each element (entry) is a key-value pair (k, v), where k is the key and v can be any object to store additional information. Each key is unique. (different entries have different keys.) Map support the following methods: Size(): Return the number of entries in M isEmpty(): Test whether M is empty get(k); If M contains an entry e with key=k, then return e else return null. put(k, v): If M does not contain an entry with key=k then add (k, v) to the map and return null; else replace the entry with (k, v) and return the old value.

2. Binary Search Trees2 Methods of Map (continued) remove(k): remove from M the netry with key=k and return its value; if M has no such entry with key=k then return null. keys(); Return an iterable collection containing all keys stored in M values(): Return an iterable collection containing all values in M entries(): return an iterable collection containing all key-value entries in M. Remakrs: hash table is an implementation of Map.

3. Binary Search Trees3 ADT for Dictionary: A Dictionary stores elements (entries). Each element (entry) is a key-value pair (k, v), where k is the key and v can be any object to store additional information. The key is NOT unique. Dictionary support the following methods: size(): Return the number of entries in D isEmpty(): Test whether D is empty find(k): If D contains an entry e with key=k, then return e else return null. findAll(k): Return an iterable collection containing all entries with key=k. insert(k, v): Insert an entry into D, returning the entry created. remove(e): remove from D an enty e, returing the removed entry or null if e was not in D. entries(): return an iterable collection of the key-value entries in D.

4. Binary Search Trees4 Part-F1 Binary Search Trees 6 9 2 4 1 8   

5. Binary Search Trees5 Search Trees Tree data structure that can be used to implement a dictionary. find(k): If D contains an entry e with key=k, then return e else return null. findAll(k): Return an iterable collection containing all entries with key=k. insert(k, v): Insert an entry into D, returning the entry created. remove(e): remove from D an enty e, returing the removed entry or null if e was not in D.

6. Binary Search Trees6 A binary search tree is a binary tree storing keys (or key-value entries) at its internal nodes and satisfying the following property: Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u)  key(v)  key(w) Different nodes can have the same key. External nodes do not store items An inorder traversal of a binary search trees visits the keys in increasing order 6 92 418

7. Binary Search Trees7 Search To search for a key k, we trace a downward path starting at the root The next node visited depends on the outcome of the comparison of k with the key of the current node If we reach a leaf, the key is not found and we return null Example: find(4): Call TreeSearch(4,root) Algorithm TreeSearch(k, v) if T.isExternal (v) return v if k  key(v) return TreeSearch(k, T.left(v)) else if k  key(v) return v else { k  key(v) } return TreeSearch(k, T.right(v)) 6 9 2 4 1 8   

8. Binary Search Trees8 Insertion To perform operation insert(k, o), we search for key k (using TreeSearch) Algorithm TreeINsert(k, x, v): Input: A search key, an associate value x and a node v of T to start with Output: a new node w in the subtree T(v) that stores the entry (k, x) W  TreeSearch(k,v) If k=key(w) then return TreeInsert(k, x, T.left(w)) T.insertAtExternal(w, (k, x)) Return Example: insert 5 Example: insert another 5? 6 92 418 6 9 2 4 18 5    w w

9. Binary Search Trees9 Deletion To perform operation remove( k ), we search for key k Assume key k is in the tree, and let v be the node storing k If node v has a leaf child w, we remove v and w from the tree with operation removeExternal( w ), which removes w and its parent and replace v with the remaining child. Example: remove 4 6 9 2 4 18 5 v w 6 9 2 5 18  

10. Binary Search Trees10 Deletion (cont.) We consider the case where the key k to be removed is stored at a node v whose children are both internal we find the internal node w that follows v in an inorder traversal we copy key(w) into node v we remove node w and its left child z (which must be a leaf) by means of operation removeExternal( z ) Example: remove 3 3 1 8 6 9 5 v w z 2 5 1 8 6 9 v 2

11. Binary Search Trees11 Deletion (Another Example) 3 1 8 6 9 4 v w z 2 4 1 8 6 9 v 2 5 5

12. Binary Search Trees12 Performance Consider a dictionary with n items implemented by means of a binary search tree of height h the space used is O(n) methods find, insert and remove take O(h) time The height h is O(n) in the worst case and O(log n) in the best case Later, we will try to keep h =O(log n). Review the past

13. Binary Search Trees13 Part-F2 AVL Trees 6 3 8 4 v z

14. Binary Search Trees14 AVL Tree Definition (§ 9.2) AVL trees are balanced. An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1. An example of an AVL tree where the heights are shown next to the nodes:

15. Binary Search Trees15 Balanced nodes A internal node is balanced if the heights of its two children differ by at most 1. Otherwise, such an internal node is unbalanced.

16. Binary Search Trees16 Height of an AVL Tree Fact: The height of an AVL tree storing n keys is O(log n). Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h. We easily see that n(1) = 1 and n(2) = 2 For n > 2, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and another of height n-2. That is, n(h) = 1 + n(h-1) + n(h-2) Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction), n(h) > 2 i n(h-2i)>2 {  h/2  -1} (1) = 2 {  h/2  -1} Solving the base case we get: n(h) > 2 h/2-1 Taking logarithms: h < 2log n(h) +2 Thus the height of an AVL tree is O(log n) 3 4 n(1) n(2) h-1 h-2

17. Binary Search Trees17 Insertion in an AVL Tree Insertion is as in a binary search tree Always done by expanding an external node. Example: 44 1778 325088 4862 54 w b=x a=y c=z 44 1778 325088 4862 before insertionafter insertion It is no longer balanced

18. Binary Search Trees18 Names of important nodes w: the newly inserted node. (insertion process follow the binary search tree method) The heights of some nodes in T might be increased after inserting a node. Those nodes must be on the path from w to the root. Other nodes are not effected. z: the first node we encounter in going up from w toward the root such that z is unbalanced. y: the child of z with higher height. y must be an ancestor of w. (why? Because z in unbalanced after inserting w) x: the child of y with higher height. x must be an ancestor of w. The height of the sibling of x is smaller than that of x. (Otherwise, the height of y cannot be increased.) See the figure in the last slide.

19. Binary Search Trees19 Algorithm restructure(x): Input: A node x of a binary search tree T that has both parent y and grand-parent z. Output: Tree T after a trinode restructuring. 1. Let (a, b, c) be the list (increasing order) of nodes x, y, and z. Let T0, T1, T2 T3 be a left-to-right (inorder) listing of the four subtrees of x, y, and z not rooted at x, y, or z. 2. Replace the subtree rooted at z with a new subtree rooted at b.. 3. Let a be the left child of b and let T0 and T1 be the left and right subtrees of a, respectively. 4. Let c be the right child of b and let T2 and T3 be the left and right subtrees of c, respectively.

20. Binary Search Trees20 Restructuring (as Single Rotations) Single Rotations:

21. Binary Search Trees21 Restructuring (as Double Rotations) double rotations:

22. Binary Search Trees22 Insertion Example, continued 88 44 17 78 3250 48 62 2 4 1 1 2 2 3 1 54 1 T 0 T 1 T 2 T 3 x y z unbalanced......balanced T 1

23. Binary Search Trees23 Theorem: One restructure operation is enough to ensure that the whole tree is balanced. Proof: Look at the four cases on slides 20 and 21.

24. Binary Search Trees24 Removal in an AVL Tree Removal begins as in a binary search tree by calling removal(k) for binary tree. may cause an imbalance. Example: 44 17 783250 88 48 62 54 44 17 7850 88 48 62 54 before deletion of 32after deletion w

25. Binary Search Trees25 Rebalancing after a Removal Let z be the first unbalanced node encountered while travelling up the tree from w. w-parent of the removed node (in terms of structure, not the name) let y be the child of z with the larger height, let x be the child of y defined as follows; If one of the children of y is taller than the other, choose x as the taller child of y. If both children of y have the same height, select x be the child of y on the same side as y (i.e., if y is the left child of z, then x is the left child of y; and if y is the right child of z then x is the right child of y.) The way to obtain x, y and z are different from insertion.

26. Binary Search Trees26 Rebalancing after a Removal We perform restructure(x) to restore balance at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 44 17 7850 88 48 62 54 c=x b=y a=z 44 17 78 5088 48 62 54 w

27. Binary Search Trees27 Unbalanced after restructuring 44 17 7850 88 62 w c=x b=y a=z 44 17 78 5088 62 32 1 1 h=5 h=3 h=4 Unbalanced balanced

28. Binary Search Trees28 Rebalancing after a Removal We perform restructure(x) to restore balance at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 44 17 7850 88 48 62 54 w c=x b=y a=z 44 17 78 5088 48 62 54

29. Binary Search Trees29 Example a: Which node is w? Let us remove node 17. 44 17 783250 88 48 62 54 44 32 7850 88 48 62 54 before deletion of 32after deletion w

30. Binary Search Trees30 Rebalancing: We perform restructure(x) to restore balance at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached 44 32 7850 88 48 62 54 c=x b=y a=z 44 32 78 5088 48 62 54 w

31. Binary Search Trees31 Running Times for AVL Trees a single restructure is O(1) using a linked-structure binary tree find is O(log n) height of tree is O(log n), no restructures needed insert is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log n) remove is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log n)