1. 1 Joe Meehean

2.  BST efficiency relies on height lookup, insert, delete: O(height) a balanced tree has the smallest height  We can balance an unbalanced tree DSW expensive  How can we ensure that our BST stays balanced? after inserts and deletes 2

3.  Adelson-Velskii and Landis  Binary search tree  Additional balance condition for every node height of subtrees differ by at most 1 must store height of each node empty tree has height of -1 3

4.  Insert as normal (BST)  Update heights  Check balance condition height differences of nodes along path from new node to root  Use rotation to fix imbalances 4

5.  Fixing imbalance at node A  4 cases: Insertion into 1. left subtree of left child of A 2. right subtree of left child of A 3. left subtree of right child of A 4. right subtree of right child of A  Cases 1 & 4 are outside inserts  Cases 2 & 3 are inside inserts 5

6.  Insert key less than A & B 6 A A B B X Y Z h: 2 h: 3h: 2 h: 4

7.  Insert key less than A & B 7 A A B B X Y Z h: 2 h: 3h: 2 h: 4 A A B B X Y Z h: 2 h: 3 h: 4h: 2 h: 5

8.  What series of rotations can we do to fix this? 8

9.  Rotate A with “highest” subtree 9 A A B B X Y Z h: 2 h: 3 h: 4h: 2 h: 5

10.  Rotate A with “highest” subtree 10 A A B B Y Z A A B B Y Z X X h: 2 h: 3 h: 4h: 2 h: 5 h: 2 h: 3 h: 4 h: 3 h: 2

11.  Are we done?  Worry about imbalances above B? 11 A A B B Y Z X h: 2 h: 3 h: 4 h: 3 h: 2

12.  B tree same height as A tree before insert  Property of outside insert rebalances  Right-right imbalance similar 12 A A B B Y Z X h: 2 h: 3 h: 4 h: 3 h: 2

13.  More specifically  Balance factor = h(left) – h(right)  If node N has a balance factor of 2, left subtree (L) is too high if L has a balance factor of 1 rotate N & L  If node N has a balance factor of -2, right subtree (R) is too high if R has a balance factor of -1 rotate N & R 13

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15.  Fixing imbalance at node A  4 cases: Insertion into 1. left subtree of left child of A 2. right subtree of left child of A 3. left subtree of right child of A 4. right subtree of right child of A  Cases 1 & 4 are outside inserts  Cases 2 & 3 are inside inserts 15

16.  Insert key less than A, greater than B 16 A A B B X Y Z h: 2 h: 3h: 2 h: 4

17.  Insert key less than A, greater than B 17 A A B B X Y Z h: 2 h: 3h: 2 h: 4 A A B B X Y Z h: 3 h: 2 h: 4h: 2 h: 5

18.  Let’s try to rotate A & B again 18 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Y

19.  Let’s try to rotate A & B again 19 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Y A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Y

20.  Let’s expand Y 20 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Q R C C A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Y h: 1 h: 2

21.  What if we made C the root 21 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 R C C h: 2 A A B B X Z h: 3 h: 1 h: 3 h: 2 h: 4 C C h: 2 Q R h: 1 Q

22.  What series of rotations can we do to accomplish this? 22

23.  Rotate left child (B) with its right child (C) 23 A A B B X Z h: 3 h: 1 h: 4h: 2 h: 5 C C h: 2 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 C C h: 2 Q R h: 1 Q R

24.  Rotate root (A) with its new left child (C) 24 A A B B X Z h: 3 h: 1 h: 3 h: 2 h: 4 C C h: 2 A A B B X Z h: 3 h: 2 h: 4h: 2 h: 5 Q R C C h: 2 Q R

25.  Are we done?  Worry about imbalances above ? 25 A A B B X Z h: 3 h: 1 h: 3 h: 2 h: 4 C C h: 2 Q R

26.  C tree same height as A tree before insert  Property of inside insert rebalances  Right-left imbalance similar 26 A A B B X Z h: 3 h: 1 h: 3 h: 2 h: 4 C C h: 2 Q R

27.  More generally  If there is an inside imbalance at A  Find A’s highest child, B  Rotate B with its highest child, C  Rotate A with C 27

28.  More specifically  Balance factor = h(left) – h(right)  If node N has a balance factor of 2, left subtree (L) is too high if L has a balance factor of -1 rotate L with L’s right child rotate N with its new left child 28

29.  More specifically  Balance factor = h(left) – h(right)  If node N has a balance factor of -2, right subtree (R) is too high if R has a balance factor of 1 rotate R with R’s left child rotate N with it’s new right child 29

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31.  Find the node N w/ key to be deleted  Different actions depending on N’s # of kids Case 1: N has 0 kids (it’s a leaf)  set parent’s N-pointer (left or right) to null Case 2: N has 1 kid  set parent’s N-pointer to point to N’s only kid Case 3: N has 2 kids  Replace N’s key with a key further down in the tree  Delete that node 31

32.  What node value can replace n’s value? new value of n must be: > all values in left subtree < all values in right subtree  Largest value from the left subtree  Smallest value from the right subtree let’s choose this one (arbitrarily) use findMin on root of right subtree 32

33.  Deletes can also imbalance an AVL tree  Can we fix these imbalances with rotations too?  How many rotations do we need to do? 33

34. 75 55 50 60 57 Smallest value in right subtree delete(50) 34 X Y Z h: 2 h: 3 h: 1 h: 0 h: 2 h: 4 65 h: 1 h: 5 62 h: 0

35. Copy smallest key in R subtree 35 75 55 60 57 delete(50) X Y Z h: 2 h: 3 h: 1 h: 0 h: 2 h: 4 65 h: 1 h: 5 62 h: 0

36. delete(55) from R subtree 36 75 57 55 60 delete(50) X Y Z h: 2 h: 3 h: 0 h: 2 h: 4 65 h: 1 h: 5 62 h: 0

37. 75 57 55 60 Fix imbalance delete(50) 37 X Y Z h: 1 h: 2 h: 0 h: 2 h: 3 65 h: 1 h: 5 62 h: 0

38. New imbalance Need to rebalance all the way up to root 38 75 57 55 60 delete(50) X Y Z h: 1 h: 2 h: 0 h: 3 65 h: 5 62 h: 0

39.  More generally  Perform a BST delete  Check for imbalances along path to root  Rebalance using expanded insert rules if delete happened in “light” child “heavy” child may have a balance factor of 0 additional case from insertion rules 39

40.  Balance factor = h(left) – h(right)  Given node N and its two children L & R  bf(N) == -2 & bf(R) == 0 single right child rotation  bf(N) == -2 & bf(R)== -1 single right child rotation  bf(N) == -2 & bf(R) == 1 double right child rotation (from inside insert) 40

41.  Balance factor = h(left) – h(right)  Given node N and its two children R & L  bf(N) == 2 & bf(L) == 0 single left child rotation  bf(N) == 2 & bf(L) == 1 single left child rotation  bf(N) == 2 & bf(L) == -1 double left child rotation (from inside insert) 41

42.  BST insert, delete, lookup O(H), H is height of tree  AVL like BST, but ensures balance balanced tree has height of log N  Balancing rotations on insert and delete 42

43.  AVL Complexity  Insert inserting node: O(H) == O(logN) rebalance: O(rotations) == O(2) == O(1)  Erase deleting a node: O(H) == O(logN) rebalance: O(rotations) == O(logN) 43

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