AVL Trees

AVL TreesAVL Trees

RecallRecall

Motivation:Motivation: Linked lists and queues work well enough for most applications, but provide slow service for large data sets.

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Ordered insertion takes too long for large sets.

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Solution: TreesSolution: Trees

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Binary Trees provide quick access to data

O (logN)vs

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O (logN)vs

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O(log N)

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Why itmatters

Why itmatters

Trees: BalanceTrees: Balance4444

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Consider animbalanced Binary Search Tree. Search performancedegrades back into a linked list.

The promisedO(log N) will approach O(N).

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BalanceBalance

If we had someway of keeping the tree “in balance”, our searches would be closer to O(log N)

But how does one measure balance?

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Measuring “Balance”Measuring “Balance”

…but the height alone does notgive us a indication of imbalance

Perhaps wemeasurethe height...

Balance: The KeyBalance: The Key

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Subtree’s rightchild too tall forits sibling

Balance is a relative property, not an absolute measure...

…as withall things.

Measuring “Balance”Measuring “Balance”Adelson-Velskii and Landis suggested that, since trees may be defined recursively, balance may also be determined recursively.

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Internal nodes are also roots of subtrees.

If each node’s subtree has a similar height, the overall tree is balanced

Measuring BalanceMeasuring Balance

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HH-1

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Thus, the actual numerical height does not determine balance; rather, it’s the relativerelative height of both subtrees.

Balanced Tree Test:Balanced Tree Test: subtrees differ in height by no more than one. This recursively holds true for all nodes!

Balance FactorBalance FactorTo measure the balance of a node, there are two steps:

Record the node height (NH)--an absolute numeric measure

Record the node’s balance factor (BF)--a relative measure

The node height alone does not indicate balance: we have to compare it to the height of other nodes.

Node / Tree HeightNode / Tree Height

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Tree height is easy, Here are three trees.

Height?Height?

H=1

H=2

H=3

STEP 1

STEP 1

Node / Tree Heights for AVLsNode / Tree Heights for AVLs

By convention, a null reference will return aheight of zero (0).

Therefore:The Balance Factor of 12121212

is Ht of minus (0)17171717

= 1 – (0) = +1

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Balance FactorBalance Factor

Remember that balance is relativerelative.

So we define the Balance Factor (BF) at:

BF = (right subtree height - left subtree height)

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STEP 2

STEP 2

Balance FactorBalance Factor

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Each node has a balance factor. Remember,it’s based on (right ht - left ht).

(We will note BF outside the node, to avoid confusion with the key values. Some texts don’t even give the keys, because values are irrelevant once you have a proper BST.)

Quiz YourselfQuiz Yourself

Calculate the balance factor for each node ineach tree:

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Quiz Yourself- AnsQuiz Yourself- AnsCalculate the balance factor for each node in each tree:

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Here, the height is written inside the node, covering the key value it holds.

Hopefully, this graphical clue helps.

Using Balance FactorsUsing Balance Factors

How do the Balance Factors help?

Recall: Adelson-Velskii and Landisused a recursive definitionof trees. Our BF measureis likewiserecursive.

If any BF is > +1 or < -1, there’s imbalance

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Balanced

NotBalanced

AVL’s “Weak” BalancingAVL’s “Weak” Balancing• A fully balanced tree guarantees that, for nodes on each level, all

left and right subtrees have the same height, and imposes this recursively.

• AVL trees use a weaker balance definition which still maintains the logarithmic depth requirement:

– for any node in an AVL tree, the height of the left and right nodes differs by at most 1.

• For example, this is a valid AVL tree, but not strictly balanced:

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Addressing ImbalanceAddressing Imbalance

If we build a tree from scratch, and check nodes for imbalance after a new insertion, we can guard against imbalance.

Q:Q: So what do we do when the tree is unbalanced?

A:A: Rotate the tree.

Identifying Offending ScenariosIdentifying Offending ScenariosThe key to fixing an unbalanced tree is knowing what caused it to break. Let’s inductively look at what can cause a good tree to go bad.

Given an AVL balanced tree:

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0Two possible left insertions could break balance:

Insertion toleft subtreeof left child

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Insertion toright subtreeof left child

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PatternsPatternsof Problemsof ProblemsInsertion to

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Insertion toright subtreeof left child

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Insertion toleft subtreeof right child

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… and the mirror of these insertions

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Insertion toright subtreeof right child

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Insertion toright subtreeof right child

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Insertion toleft subtreeof right child

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Insertion toright subtreeof left child

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Given our recursive definition of trees, we know these four scenarios are complete. The same insertion into a larger AVL balanced tree does not create a new category.

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Imbalance foundImbalance foundregardless of heightregardless of height

matchesmatches

(H-1) - (H+1) = -2

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Insertion toright subtreeof right child

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Insertion toleft subtreeof right child

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Insertion toright subtreeof left child

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Let’s look at two Let’s look at two insertions thatinsertions that

cause imbalance. cause imbalance. Recall they are Recall they are

mirrors.mirrors.

They both deal with They both deal with “outside” insertions “outside” insertions on the tree’s exterior on the tree’s exterior face: the left-left and face: the left-left and right-right insertions.right-right insertions.

Single RotationSingle Rotation

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Insertion toright subtreeof right child

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-2#1These insertions can be cured by swapping parent and child . . .

. . . provided you maintain search order. (An AVL tree is a BST)

Single Rotation: RightSingle Rotation: Right

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A single rotation round the A single rotation round the unbalanced node restores AVL unbalanced node restores AVL balance in all casesbalance in all cases

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Think of rotations - Imagine the joints at the ‘4’ node starting to flex and articulate.

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A similar rotation is used for right-rightinsertions that cause an imbalance

Insertion toright subtreeof right child

Single Rotation: LeftSingle Rotation: Left

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Word of cautionWord of caution

The single rotations are simple, but you have to keep track of child references.

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Insertion toright subtreeof right child

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Insertion toleft subtreeof right child

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Insertion toright subtreeof left child

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Let’s look at theLet’s look at theother two insertionsother two insertionscausing problems. causing problems.

Recall they are Recall they are mirrors.mirrors.

They both deal with They both deal with “inside” insertions on “inside” insertions on

the tree’s interior:the tree’s interior:right-left and left-rightright-left and left-right

Insertion toright subtreeof left child

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Insertion toleft subtreeof right child

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Double RotationDouble Rotation

These insertions are not solved with a single rotation.

Instead, two single rotations, left-right and right-left are performed.

Why Doesn’t Single Rotation Work?Why Doesn’t Single Rotation Work?

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RotationRotationAttemptedAttempted

Unbalanced insertions into the “interior” of the tree remain unbalanced with only a single rotation

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Double RotationDouble Rotation

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(trivial case)

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Insertion toright subtreeof left child

Double RotationDouble Rotation

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MagicMagic

Step?Step?

Double RotationDouble Rotation

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Solved: Move up X! W becomes X’s left child Y becomes X’s right child W’s right child becomes X’s left child X’s right child (if any) becomes Y’s left child

Double Rotation StrategyDouble Rotation Strategy

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SLRSLR

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On a left-right insert, we first rotate left

This places the tree in a position for which we already have a working solution. (Code reuse!)

Double left rotate =Double left rotate =single left rot + single right rotsingle left rot + single right rot

Double Rotation StrategyDouble Rotation Strategy

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Similarly, we perform a right and left rotation oninsertions that went left-right. The first rotation reforms the tree into an exterior imbalance. We already have a way of fixing this: single left rotations