CSC 213

Lecture 9: Red-Black Trees

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To check if your compiler from lab is working, you see what tokens your system outputs using this command:

java –cp java_cup.jar;. mpc filename

This will generate a file filename.token which contains all the tokens you found.

(2,4) Trees: Pro & Con

Pros: (2,4) Trees are balanced trees with no

rotations Contains many fewer balancing cases

than AVL or splay trees

Cons: Cannot use any BST code we already

wrote

&%#$*#*: n-node name is a crime and should be

abolished

Red-Black TreesBinary tree representation of a (2,4) tree

Mark nodes red when entries share a node in (2,4) tree Mark nodes black when also parent & child in (2,4) tree

Get to use much of our existing BST code!

2 6 73 54

4 6

2 7

5

3

3

5OR

Red-Black Trees (§ 9.5)

Draw the equivalent (2,4) tree:

9

154

62 12

7

21

Red-Black Trees (§ 9.5)Red-black trees are BSTs satisfying these properties:

Root Property: Root node is black External Property: All leaf nodes are black Internal Property: Red nodes only have black children Depth Property: All leaves have same black depth

Black depth = number of black ancestors a node has9

154

62 12

7

21

Height of a Red-Black TreeRed-black tree storing n entries has height O(log n)

Remember, it is functionally equivalent to (2,4) tree

Search a red-black tree identical to searching a BST

Searching red-black tree takes ______________ time

Insertioninsert(k, o):

Perform usual BST insertion and color external nodes black If new node, z, is root, then color it black, else color z red If parent, v, of z is black, internal property is preserved Else (v is red) we have double red and must reorganize tree

Example: insert(3) is all good

6

8

6

3 8z

v v

z

Insertioninsert(k, o):

Perform usual BST insertion and color external nodes black If new node, z, is root, then color it black, else color z red If parent, v, of z is black, internal property is preserved Else (v is red) we have double red and must reorganize tree

Example: insert(4) causes double red

6

3 8

6

3 8

4z

v v

z

Remedying a Double RedDouble red with child z, parent v, and aunt, wCase 1: w is red Double red is a 5-node; recoloring is split equivalent

4

6

7z

v

2 4 6 7

2w

RecoloringRecoloring remedies double red when uncle is redMake v and w black and grandparent, u, red

If u is root, however, it must stay black

Equivalent to splitting a 5-node Just like in (2,4) tree, double red may propagate up to u

2 4 6 76 7

… 4 …

2

4

6

7z

v2

w 4

6

7z

v2

w

Remedying a Double RedDouble red with child z, parent v, and aunt, wCase 2: w is black Double red is illegal structure of a legal 4-node

4

6

7z

vw2

4 6 7

.. 2 ..

RestructuringRestructuring remedies double red when uncle is blackEquivalent (2,4) tree never changes

This only fixes an error in red-black tree’s organizationSince only reorganizes red-black tree, cannot propagate further

4

6

7z

vw2

4 6 7

.. 2 ..

4

6

7

z

v

w2

4 6 7

.. 2 ..

Restructuring (cont.)Four ways we perform this restructuring Just depends on whether double red nodes

are left or right childrenResult is always the same!

4

6

7

7

4

6

7

6

4

4

7

6

4 7

6

Analysis of InsertionTree has O(log n) height

Step 1 takes ______ time

Step 2 takes ______ time

Recoloring takes _____ time

Restructuring takes ____ time

Step 3 takes _______ time

Insertion takes ______ time!

Algorithm insert(k, o)

1. Search for k to find insert node z

2. Add new entry at node z and color z red

3. while isRed(z) && isRed(parent(z))if isBlack(sibling(parent(z)))

z restructure(z)return

else /* isRed(sibling(parent(z)) */ recolor(z) z parent(z)

Deletionremove(k) starts by performing normal BST deletionRemove Entry at node v, with w being external node removed, and r is its sibling

If either v or r was red, color r black

Example: remove(1)6

3 8

4v

r w

1

6

3 8

4

r

r

Deletionremove(k) starts by performing normal BST deletionRemove Entry at node v, with w being external node removed, and r is its sibling

If either v or r was red, color r black Else (v and r were both black), color r double black

This violates the internal property

Example: remove(8) causes double black:6

3 8

4

v

r w

6

3

4

r

Remedying a Double Black

Different remedy for double black depending on state of sibling, yCase 1: y is black and has a red child Just need to restructure (from above) tree

Case 2: y is black and children are black Underflow in (2,4) tree; perform recoloring

Case 3: y is red Use adjustment to represent 3-node better After adjustment, apply case 1 or case 2

(2,4) Tree Transfer

Restructuring double black like performing transfer in (2,4) tree

9

6 8 10

8

6 9

9

6 10

8

8

6 9

(2,4) Tree Fusion

Recoloring double black like performing fusion in (2,4) tree

5 9

6 10

5

… 6 9

5

10

9

6

…

…

5

9

6

…

(2,4) Tree Fusion

Recoloring double black like performing fusion in (2,4) tree

4

1 7 1 4

4

1 7

4

1

44

Adjustment

Adjusting double black lets us determine whether we need to recolor or restructure

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5 10

6

9

5

6

Red-Black Tree Reorganization

Insertion remedy double red

Red-black tree action (2,4) tree action result

restructuringchange of 4-node representation

double red removed

recoloring splitdouble red removed or propagated up

Red-Black Tree Reorganization

Deletion remedy double black

Red-black tree action (2,4) tree action result

restructuring transferdouble black removed

recoloring fusiondouble black removed or propagated up

adjustmentchange of 3-node representation

restructuring or recoloring follows

Your Turn

Insert 1, 2, 3, 4, 5, 6, 7, 8 into a red-black treeThen delete 3, 6, 5, 2, 8, 4, 1, 7 from your tree

Daily Quiz

Write node class for Red-Black tree