CS 240 Fall 2015 Mike Lam, Professor I find your lack of balance - - PowerPoint PPT Presentation

CS 240 Fall 2015

Mike Lam, Professor

Balanced (AVL) Trees

I find your lack of balance disturbing.

Review

• Binary Search Trees (BSTs)

– Ordered binary tree – Operations are O(h) where h is the height of the tree – For mostly-random insertions and deletions, h ≈ log n

• As in the last part of Monday's lab

– For other situations, we need to use a more "balanced"

binary tree implementation

• General idea: restore balance after every add or remove
• However, this will take extra work!
• Tradeoff between cost and benefit of balancing

Issues

• How much should we re-balance?

– How often? How many nodes? How strict? – How do we measure balance?

• We could re-balance the entire tree after every insertion

– This would lead to O(n log n) insertion times – Essentially re-build the tree every time – This seems quite excessive!

• Goal: faster insertions and "good enough" balance

– AVL trees (easiest to understand) – Red-Black trees – Many others...

AVL Trees

• AVL Trees

– Inventors G. M. Adelson-Velsky and E. M. Landis (1962) – Balance factor: height(node->left) – height(node->right) – Height-balance property

• Heights of children differ by at most one
• Balance factor is -1, 0, or 1

– Enforce this property using tree rotations

Image from: http://www.geeksforgeeks.org/how-to-determine-if-a-binary-tree-is-balanced/

Tree Height

• Height: # of edges from root to furthest leaf

– Minor extension: height of an empty tree will be -1 – Applies to whole trees as well as subtrees

Caveat: Heights can't really be negative (what would it mean to have negative edges?), but it does make the math work better...

Binary Search Tree

44 17 8 32 28 29 88 65 97 93 54 82 76 80

Binary Search Tree

44 17 8 32 28 29 88 65 97 93 54 82 76 80 h=0 h=0 h=0 h=0 h=0

Binary Search Tree

44 17 8 32 28 29 88 65 97 93 54 82 76 80 h=0 h=0 h=0 h=0 h=0 h=1 h=1 h=1

Binary Search Tree

44 17 8 32 28 29 88 65 97 93 54 82 76 80 h=0 h=0 h=0 h=0 h=0 h=1 h=1 h=1 h=-1 h=-1

Binary Search Tree

44 17 8 32 28 29 88 65 97 93 54 82 76 80 h=0 h=0 h=0 h=0 h=0 h=1 h=1 h=1

Imbalances

• How do we fix imbalances?

– Need to re-arrange tree

• Solution: Rotations!
• Example:

Imbalances

• How do we fix imbalances?

– Need to re-arrange tree

• Solution: Rotations!
• Example:

Rotations

• Rotations

– Single rotation (below) – Single/double rotations (right)

• Four cases of trinode restructuring

x x y y

x->right = y->left y->left = x y->left = x->right x->right = y

"right rotation" "left rotation"

AVL Trees

• Insertion

– Insert into BST as usual

• Binary search until empty subtree is found, then add a new node

– Check ancestors of new node for imbalances – Fix imbalances via trinode restructuring

• Removal

– Remove from BST as usual – Check ancestors of removed node for imbalances – Fix imbalances via trinode restructuring

AVL Tree

44 17 32 78 50 88 48 62

Insert 54

h=0 h=0 h=0 h=0 h=1 h=1 h=2 h=3

AVL Tree

44 17 32 78 50 88 48 62

Update heights of ancestors and check for imbalances

h=0 h=0 h=0 h=0 h=1 h=1 h=2 h=3 54 h=0

AVL Tree

44 17 32 78 50 88 48 62

Update heights of ancestors and check for imbalances

h=0 h=1 h=0 h=0 h=1 h=1 h=2 h=3 54 h=0

AVL Tree

44 17 32 78 50 88 48 62

Update heights of ancestors and check for imbalances

h=0 h=1 h=0 h=0 h=2 h=1 h=2 h=3 54 h=0

AVL Tree

44 17 32 78 50 88 48 62

Imbalance detected

h=0 h=1 h=0 h=0 h=2 h=1 h=3 h=4 54 h=0

AVL Tree

44 17 32 78 50 88 48 62

Nodes involved

h=0 h=1 h=0 h=0 h=2 h=1 h=3 h=4 54 h=0

AVL Tree

44 17 32 78 50 88 48 62

Rebalance

h=0 h=1 h=0 h=0 h=2 h=1 h=3 h=4 54 h=0

AVL Tree

44 17 32 78 62 88 50 54

Rebalance

h=1 h=0 h=0 h=0 h=2 h=1 h=3 h=4 48 h=0

AVL Tree

44 17 32 62 50 78 48 54

Balanced subtree; continue up tree to root

h=0 h=0 h=0 h=1 h=1 h=1 h=2 h=3 88 h=0

AVL Tree

44 17 32 62 50 78 48 54

Done!

h=0 h=0 h=0 h=1 h=1 h=1 h=2 h=3 88 h=0

Exercises

• 3, 9, 12, 5, 4, 1
• 10, 5, 7, 15, 9, 25
• Show every step!

– Mark rotations as L or R

To check your answers:

http://www.qmatica.com/DataStructures/Trees/AVL/AVLTree.html

AVL Tree Analysis

• In general: n(h) = 1 + n(h-1) + n(h-2)

– n(h) = number of nodes in AVL tree of height h – Reasoning: AVL tree with minimal number of nodes has one

node and two subtrees: one with height h-1 and one with height h-2

• (They differ by at most one b/c of the balance property, and you can't

skip heights)

• This is a Fibonacci progression

– n(h) >= Fib(h) = (Φ = golden ratio) – Exponential w.r.t. height: n(h) is Ω(Φh) – Thus, h is O(log n)

⌊

Φ

h

√ 5⌋

Alternative: Red-Black Trees

• Coloring scheme

– Root is colored black – All children of a red node must be colored black (no "double reds") – All nodes with zero or one children have the same number of black-colored ancestors

• Path from root to furthest leaf is no more than twice as long as the path from

root to nearest leaf

• Less-strictly balanced
• Faster insertion/removal but slower lookups