1. Binary trees -2 Chapter 6 1

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3. Threaded trees (depth first) Binary trees have a lot of wasted space: the leaf nodes each have 2 null pointers We can use these pointers to help us in inorder traversals We have the pointers reference the next node in an inorder traversal; called threads We need to know if a pointer is an actual link or a thread, so we keep a Boolean for each pointer 3

4. Threaded Tree Example 4 8 75 3 11 13 1 6 9

5. 5 Threaded Tree Traversal Start at the leftmost node in the tree, print it, and follow its right thread – following a thread to the right, will output the node and continue to its right – following a link to the right, will go to the leftmost node, print it, and continue

6. 6 Threaded Tree Traversal -1 8 75 3 11 13 1 6 9 Start at leftmost node, print it Output 1

7. 7 Threaded Tree Traversal -2 8 75 3 11 13 1 6 9 Follow thread to right, print node Output 1 3

8. 8 Threaded Tree Traversal -3 8 75 3 11 13 1 6 9 Follow link to right, go to leftmost node and print Output 1 3 5

9. 9 Threaded Tree Traversal -4 8 75 3 11 13 1 6 9 Follow thread to right, print node Output 1 3 5 6

10. 10 Threaded Tree Traversal -5 8 75 3 11 13 1 6 9 Follow link to right, go to leftmost node and print Output 1 3 5 6 7

11. 11 Threaded Tree Traversal - 6 8 75 3 11 13 1 6 9 Follow thread to right, print node Output 1 3 5 6 7 8

12. Amir Kamil8/8/0212 Threaded Tree Traversal - 7 8 75 3 11 13 1 6 9 Follow link to right, go to leftmost node and print Output 1 3 5 6 7 8 9

13. 13 Threaded Tree Traversal -8 8 75 3 11 13 1 6 9 Follow thread to right, print node Output 1 3 5 6 7 8 9 11

14. 14 Threaded Tree Traversal -9 8 75 3 11 13 1 6 9 Follow link to right, go to leftmost node and print Output 1 3 5 6 7 8 9 11 13

15. Threaded Tree Traversal Code 15 Node leftMost(Node n) { Node ans = n; if (ans == null) { return null; } while (ans.left != null) { ans = ans.left; } return ans; } void inOrder(Node n) { Node cur = leftmost(n); while (cur != null) { print(cur); if (cur.rightThread) { cur = cur.right; } else { cur = leftmost(cur.right); }

16. Traversal Through Tree Transformation (read only) Systematic visiting of all nodes in the tree. Each node visited once, does not specify order. Breadth-first traversal: visit each node, starting from the lowest level and moving down by level, visiting nodes from left to right. Depth-first traversals: go in subtree as deep as you can, backtrack. Differ on the order of visiting root, left, right subtrees. preorder: VLR. inorder: LVR. postorder: LRV. 16

17. Insertion searching does not modify the tree. To insert a new node with key el (new), a tree node with a dead end has to be reached, new node attached to it. found using same procedure as searching: compare key of currently scanned node el (cur) to el (new). If el (new) less than the key el (cur) try left child; otherwise try right child. If the child is empty, discontinue search and make the child point to a new node of key el. 17

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19. Inserting in threaded tree 19 for inorder, only takes care of successors. Node with a right child: has its successor somewhere in the right subtree, does not need a thread. Why ? Threads are for "climbing up the tree", not for going down. A node with no right child has its successor somewhere. Inherits successor from parent. If a node becomes a left node, its parent is successor.

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21. Deleting (try) 21

22. Balancing a tree The main idea of DSW algorithm is a rotation There are two types of rotation, left and right Rotate Right( Gr, Par, Ch ) – If Par is not the root of the tree Grandparent Gr of child Ch, becomes Ch’s parent by replacing Par; – Right subtree of Ch becomes left subtree of Ch’s parent Par; – Node Ch acquires Par as its right child 22

23. A picture will help 23

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26. AVL tree The height of left and right subtree of every node differ by at most one. 26

27. AVL Trees27 Balanced and unbalanced BST 4 25 13 1 5 2 4 3 7 6 4 26 5713 Is this “balanced”?

28. AVL Trees28 Perfect Balance Want a complete tree after every operation – tree is full except possibly in the lower right This is expensive – For example, insert 2 in the tree on the left and then rebuild as a complete tree Insert 2 & complete tree 6 49 815 5 28 6914

29. AVL Trees29 AVL - Good but not Perfect Balance AVL trees are height-balanced binary search trees Balance factor of a node – height(left subtree) - height(right subtree) An AVL tree has balance factor calculated at every node – For every node, heights of left and right subtree can differ by no more than 1 – Store current heights in each node

30. AVL Trees30 Node Heights 1 00 2 0 6 49 815 1 height of node = h balance factor = h left -h right empty height = -1 0 0 height=2 BF=1-0=1 0 6 49 15 1 Tree A (AVL)Tree B (AVL)

31. AVL Trees31 Node Heights after Insert 7 2 10 3 0 6 49 815 1 height of node = h balance factor = h left -h right empty height = -1 1 0 2 0 6 49 15 1 0 7 0 7 balance factor 1-(-1) = 2 Tree A (AVL)Tree B (not AVL)

32. AVL Trees32 Insert and Rotation in AVL Trees Insert operation may cause balance factor to become 2 or –2 for some node – only nodes on the path from insertion point to root node have possibly changed in height – So after the Insert, go back up to the root node by node, updating heights – If a new balance factor (the difference h left -h right ) is 2 or –2, adjust tree by rotation around the node

33. AVL Trees33 Single Rotation in an AVL Tree 2 10 2 0 6 49 815 1 0 7 0 1 0 2 0 6 4 9 8 15 1 0 7

34. AVL Trees34 Let the node that needs rebalancing be . There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of . The rebalancing is performed through four separate rotation algorithms. Insertions in AVL Trees

35. AVL Trees35 j k XY Z Consider a valid AVL subtree AVL Insertion: Outside Case h h h

36. AVL Trees36 j k X Y Z Inserting into X destroys the AVL property at node j AVL Insertion: Outside Case h h+1h

37. AVL Trees37 j k X Y Z Do a “right rotation” AVL Insertion: Outside Case h h+1h

38. AVL Trees38 j k X Y Z Do a “right rotation” Single right rotation h h+1h

39. AVL Trees39 j k X Y Z “Right rotation” done! (“Left rotation” is mirror symmetric) Outside Case Completed AVL property has been restored! h h+1 h

40. AVL Trees40 j k XY Z AVL Insertion: Inside Case Consider a valid AVL subtree h h h

41. AVL Trees41 Inserting into Y destroys the AVL property at node j j k X Y Z AVL Insertion: Inside Case Does “right rotation” restore balance? h h+1h

42. AVL Trees42 j k X Y Z “Right rotation” does not restore balance… now k is out of balance AVL Insertion: Inside Case h h+1 h

43. AVL Trees43 Consider the structure of subtree Y… j k X Y Z AVL Insertion: Inside Case h h+1h

44. AVL Trees44 j k X V Z W i Y = node i and subtrees V and W AVL Insertion: Inside Case h h+1h h or h-1

45. AVL Trees45 j k X V Z W i AVL Insertion: Inside Case We will do a left-right “double rotation”...

46. AVL Trees46 j k X V Z W i Double rotation : first rotation left rotation complete

47. AVL Trees47 j k X V Z W i Double rotation : second rotation Now do a right rotation

48. AVL Trees48 j k X V Z W i Double rotation : second rotation right rotation complete Balance has been restored h h h or h-1

49. AVL Trees49 Implementation balance (1,0,-1) key right left No need to keep the height; just the difference in height, i.e. the balance factor; this has to be modified on the path of insertion even if you don’t perform rotations Once you have performed a rotation (single or double) you won’t need to go back up the tree

50. AVL Trees50 Example of Insertions in an AVL Tree 1 0 2 20 1030 25 0 35 0 Insert 5, 40

51. AVL Trees51 Example of Insertions in an AVL Tree 1 0 2 20 1030 25 1 35 0 5 0 20 1030 25 1 355 40 0 0 0 1 2 3 Now Insert 45

52. AVL Trees52 Single rotation (outside case) 2 0 3 20 1030 25 1 35 2 5 0 20 1030 25 1 405 0 0 0 1 2 3 45 Imbalance 35 45 00 1 Now Insert 34

53. AVL Trees53 Double rotation (inside case) 3 0 3 20 1030 25 1 40 2 5 0 20 1035 30 1 405 45 0 1 2 3 Imbalance 45 0 1 Insertion of 34 35 34 0 0 1 2534 0

54. Heap A heap is a binary tree T that stores a key- element pairs at its internal nodes. Properties : Min Heap: key(parent) < key(child) Max Heap: key(parent) > key(child)] 54

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56. Example of a complete binary max- heap if B is a child node of A, then key(A) ≥ key(B). This implies that an element with the greatest key is always in the root node, and so such a heap is sometimes called a max-heap. Alternatively, if the comparison is reversed, the smallest element is always in the root node, which results in a min-heap. 56

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60. Max heap Terminate heap when – reach root – key parent is greater than key child 60

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65. 6.8 and 6.10 no need 65