AVL trees - cs.toronto.edutabrown/csc263/2014W/week3.pdf · AVL trees. Today •AVL delete and subsequent rotations •Testing your knowledge with interactive demos! AVL tree •Is

AVL trees

Today

• AVL delete and subsequent rotations

• Testing your knowledge with interactive

demos!

AVL tree

• Is a binary search tree

• Has an additional height constraint:

– For each node x in the tree, Height(x.left) differs

from Height(x.right) by at most 1

• I promise:

– If you satisfy the height constraint, then the

height of the tree is O(lg n).

– (Proof is easy, but no time! =])

AVL tree

• To be an AVL tree, must always:

– (1) Be a binary search tree

– (2) Satisfy the height constraint

• Suppose we start with an AVL tree, then delete as if we’re in a regular BST.

• Will the tree be an AVL tree after the delete?

– (1) It will still be a BST…

– (2) Will it satisfy the height constraint?

• (Not covering insert, since you already did in class)

BST Delete breaks an AVL tree

7

4

3

9

7

4

3

Delete(9)

h(left) > h(right)+1

so NOT an AVL tree!

Replacing the height constraint

with balance factors

• Instead of thinking about the heights of nodes, it

is helpful to think in terms of balance factors

• The balance factor bf(x) = h(x.right) – h(x.left)

– bf(x) values -1, 0, and 1 are allowed

– If bf(x) < -1 or bf(x) > 1 then tree is NOT AVL

Same example with bf(x), not h(x)

7

4

3

9

7

4

3

Delete(9)

-1

-1

0

0

-2

-1

0

bf < -1

so NOT an AVL tree!

What else can BST Delete break?

• Balance factors of ancestors…

7

4

3

9

7

4Delete(3)

-1

-1

0

0

-1

-1 900

0

Need a new Delete algorithm

• Goal: if tree is AVL before Delete, then tree is AVL after Delete.

• Step 1: do BST delete.

– This maintains the BST property, but cancause the balance factors of ancestors to be outdated!

• Step 2: fix the height constraint and update balance factors.

– Update any invalid balance factors affected by delete.

– After updating them, they can be < -1 or > 1.

– Do rotations to fix any balance factors that are too small or large while maintaining the BST property.

– Rotations can cause balance factors to be outdated also!

Bad balance factors

• Start with an AVL tree, then do a BST Delete.

• What bad values can bf(x) take on?

– Delete can reduce a subtree’s height by 1.

– So, it might increase or decrease h(x.right) –

h(x.left) by 1.

– So, bf(x) might increase or decrease by 1.

– This means:

• if bf(x) = 1 before Delete, it might become 2. BAD.

• If bf(x) = -1 before Delete, it might become -2. BAD.

• If bf(x) = 0 before Delete, then it is still -1, 0 or 1. OK.

2 cases

Problematic cases for Delete(a)

• bf(x) = -2 is just symmetric to bf(x) = 2.

• So, we just look at bf(x) = 2.

x2

h+2

h

x-2

a a(deleted)

Delete(a): 3 subcases for bf(x)=2

• Since tree was AVL before, bf(z) = -1, 0 or 1

Case bf(z) = -1 Case bf(z) = 0 Case bf(z) = 1

h+1

x

T2

T1

z

2

0

T3

ah

x

T2

T1

z

2

1

T3

a

x

T2

T1

z

2

h+1

-1h

a

T3

Fixing case bf(x) = 2, bf(z) = 0

• We do a single left rotation

• Preserves the BST property, and fixes bf(x) = 2

x

T2

T1

z

2

h+1

h0

T3

z

T2

T1

x

-1

1

T3

Fixing case bf(x) = 2, bf(z) = 1

• We do a single left rotation (same as last case)

• Preserves the BST property, and fixes bf(x) = 2

x

T2

T1

z

2

h+1

h1

T3

z

T2T1

x

0

0

T3

h

Delete(a): bf(x)=2, bf(z)=-1 subcases

Case bf(z) = -1: we have 3 subcases. (More details)

Case bf(y) = 0 Case bf(y) = -1 Case bf(y) = 1

x

T1

z

2

h

h-1

T3

a y

T21 T22

h-1

0

x

T1

z

2

-1

a y

T21

22T22h

-1

x

T1

z

2

-1

a y

21T21

T22

1

T3 T3

Double right-left rotation

• All three subcases of bf(x)=2, bf(z)=-1 simply

perform a double right-left rotation.

x

z

y

y

zx

x

y

z

Delete subcases for bf(x)=2, bf(z)=-1

• Case bf(y)=0: double right-left rotation!

x

T1

z

2

h

h-1

T3

y

T21 T22

0

h

y

T1

x

0

0

T3

z

T21T22

0

Delete subcases for bf(x)=2, bf(z)=-1

• Case bf(y)=-1: double right-left rotation!

x

T1

z

2

h

h-1

T3

y

T21

-1

h

y

T1

x

0

1

T3

z

T21

22T22

0

22T22

h-1

Delete subcases for bf(x)=2, bf(z)=-1

• Case bf(y)=1: double right-left rotation!

x

T1

z

2

h

h-1

T3

y

T22

1

h

y

T1

x

0

0

T3

z

21T21

T22

-1

21T21

h-1

Recursively fixing balance factors

• Idea: start at the node we deleted, fix a

problem, then recurse up the tree to the root.

• At each node x, we update the balance factor:

bf(x) := h(bf.right) - h(bf.left).

• If bf(x) = -2 or +2, we perform a rotation.

• Then, we update the balance factors of every

node that was changed by the rotation.

• Finally, we recurse one node higher up.

Interactive AVL Deletes

• Interactive web applet