AVL Trees. Properties Insertion. October 17, 2017 Cinda Heeren / Geoffrey Tien 1

Transcription

1 AVL Trees Properties Insertion October 17, 2017 Cinda Heeren / Geoffrey Tien 1

2 BST rotation g g e h b h b f i a e i a c c f d d Rotations preserve the in-order BST property, while altering the tree structure we can use rotations to bring imbalanced trees back into balance October 12, 2017 Cinda Heeren / Geoffrey Tien 2

3 AVL trees An AVL tree is a balanced BST Each node's left and right subtrees differ in height by at most 1 Rebalancing via rotations occurs when an insertion or removal causes excessive height difference AVL tree nodes contain extra information to support this height information October 17, 2017 Cinda Heeren / Geoffrey Tien 3

4 AVL nodes enum balance_type {LEFT_HEAVY = -1, BALANCED = 0, RIGHT_HEAVY = +1}; class AVLNode { public: int data; // or template type AVLNode left; AVLNode right; balance_type balance; } AVLNode(int value) {... } AVLNode(int val, AVLNode left1, AVLNode right1) {... } AVLNode is almost the same as a binary tree node additional balance field indicates that state of subtree balance at that node October 17, 2017 Cinda Heeren / Geoffrey Tien 4

5 AVL imbalance Balanced trees: Imbalanced trees: October 17, 2017 Cinda Heeren / Geoffrey Tien 5 24

6 AVL imbalance 4 cases of imbalance C C A A B A B C A B C B LL imbalance LR imbalance RR imbalance RL imbalance Solve with a right rotation around C Left rotation around A, becomes LL case Symmetric to left imbalance cases October 17, 2017 Cinda Heeren / Geoffrey Tien 6

7 AVL insertion Maintaining balance The best way to keep a tree balanced, is to never let it become imbalanced! AVL insertion begins with ordinary BST insertion (i.e. a leaf node) followed by rotations to maintain balance i.e. AVL properties are satisfied before and after insertion if the balance attribute of a subtree's root node becomes critical (-2 or +2) as a result of inserting the new leaf, rebalance it! October 17, 2017 Cinda Heeren / Geoffrey Tien 7

8 AVL insertion Pseudocode if root is NULL Create new node containing item, assign root to it, and return true else if item is equal to root->data item exists already, return false else if item < root->data Recursively insert the item into the left subtree if height of left subtree has increased (increase variable is true) balance--; if balance == 0, reset increase variable to false if balance < -1 reset increase variable to false perform rebalanceleft else if item > root->data (symmetric to left subtree case, incrementing balance) rebalanceleft and rebalanceright are the rotations to correct the 4 imbalance cases October 17, 2017 Cinda Heeren / Geoffrey Tien 8

9 AVL insertion example Insert(65) October 17, 2017 Cinda Heeren / Geoffrey Tien 9

10 AVL insertion example Insert(65) Insert(82) 47 RR imbalance OK OK 82 OK October 17, 2017 Cinda Heeren / Geoffrey Tien 10

11 AVL insertion example Insert(65) Insert(82) Insert(87) LR imbalance OK 87 OK October 17, 2017 Cinda Heeren / Geoffrey Tien 11

12 AVL insertion example Insert(65) Insert(82) Insert(87) 71 OK OK October 17, 2017 Cinda Heeren / Geoffrey Tien 12

13 AVL tree height The height of an AVL tree containing n key values is O log n Intuition: For a fixed height h, a tree containing fewer nodes has a larger heightto-node ratio n = 15 n = 4 h = 3 = log n = O(log n) h = 3 = n 1 = O n Attempt to achieve the worst ratio by making an AVL tree of height h with the minimum number of nodes October 17, 2017 Cinda Heeren / Geoffrey Tien 13

14 AVL tree height Theorem: The height of an AVL tree with n nodes is O log n Proof: Let N h represent the minimum number of nodes in an AVL tree of height h Since the AVL property must be satisfied at every node, the children of such a tree must also be minimal, and the height difference between the children must be 1 Thus N h = 1 + N h 1 + N h 2 1 N h-1 N h-2 October 17, 2017 Cinda Heeren / Geoffrey Tien 14

15 AVL tree height N h = 1 + N h 1 + N h 2 N h 1 = 1 + N h 2 + N h 3 N h = N h 2 + N h 3 + N h 2 = 2 + 2N h 2 + N h 3 > 2N h 2 N h 2 = 1 + N h 3 + N h 4 N h > 2 2N h 4 = 2 + 2N h 4 + N h 5 > 2N h 4 N h > 2 2 2N h 6 N h > N h 8 How many times can we subtract 2 from h before we reach 0? h 2 times. N h > 2 h 2 h < log N h h O log n October 17, 2017 Cinda Heeren / Geoffrey Tien 15

16 Exercise Draw the AVL tree resulting from the following sequence of insertions: 71, 13, 65, 22, 49, 37, 45 Note the height recorded/updated at each node What would an ordinary BST look like with the same sequence of insertions? October 17, 2017 Cinda Heeren / Geoffrey Tien 16

17 Readings for this lesson Koffman Chapter 11.2 (AVL trees) AVL visualisations: (AVL tree) Next class: AVL removal October 17, 2017 Cinda Heeren / Geoffrey Tien 17