Red-Black Trees

Definitionsand

Bottom-Up Insertion

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Red-Black Trees• Definition: A red-black tree is a binary search

tree in which:– Every node is colored either Red or Black.– Each NULL pointer is considered to be a Black “node”.– If a node is Red, then both of its children are Black.– Every path from a node to a NULL contains the same number

of Black nodes.– By convention, the root is Black

• Definition: The black-height of a node, X, in a red-black tree is the number of Black nodes on any path to a NULL, not counting X.

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A Red-Black Tree with NULLs shown

Black-Height of the tree (the root) = 3Black-Height of node “X” = 2

X

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A Red-Black Tree with

Black-Height = 3

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Black Height of the tree?

Black Height of X?

X

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Bottom –Up Insertion

• Insert node as usual in BST• Color the node Red• What Red-Black property may be violated?– Every node is Red or Black?– NULLs are Black?– If node is Red, both children must be Black?– Every path from node to descendant NULL must

contain the same number of Blacks?

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Bottom Up Insertion• Insert node; Color it Red; X is pointer to it• Cases

0: X is the root -- color it Black1: Both parent and uncle are Red -- color parent and uncle

Black, color grandparent Red. Point X to grandparent and check new situation.

2 (zig-zag): Parent is Red, but uncle is Black. X and its parent are opposite type children -- color grandparent Red, color X Black, rotate left(right) on parent, rotate right(left) on grandparent

3 (zig-zig): Parent is Red, but uncle is Black. X and its parent are both left (right) children -- color parent Black, color grandparent Red, rotate right(left) on grandparent

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X

P

G

U

P

G

U

Case 1 – U is Red

Just Recolor and move up

X

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X

P

G

U

S X

P G

SU

Case 2 – Zig-Zag

Double Rotate X around P; X around G

Recolor G and X

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X

P

G

U

S P

X G

S U

Case 3 – Zig-Zig

Single Rotate P around G

Recolor P and G

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11

14

152

1 7

5 8

Black node Red node

Insert 4 into this R-B Tree

Possible insertion configurations

X (Red or Black)

Y Z

If a new node is inserted as a child of Y or Z, there is no problem since the new node’s parent is black

Possible insertion configurations

X

Y Z

If new node is child of Z, no problem since Z is black.

If new node is child of Y, no problem since the new node’s uncle (Z) is black – do a few rotations and recolor…. done

Possible insertion configurations

X

Y Z

If new node is inserted as child of Y or Z, it’s uncle will be red and we will have to go back up the tree. This is the only case we need to avoid.

Top-Down Traversal

X

Y Z

As we traverse down the tree and encounter this case, we recolor and possible do some rotations.

There are 3 cases.

Remember the goal – to create an insertion point at which the parent of the new node is Black, or the uncle of the new node is black.

Case 1 – X’s Parent is Black

X

ZY

P

X

Z

P

Just recolor and continue down the tree

Y

Case 2• X’s Parent is Red (so Grandparent is Black)

and X and P are both left/right children– Rotate P around G– Color P black– Color G red

• Note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

Case 2 diagrams

X

ZY

P

G

U

SX

ZY

P

G

US

Rotate P around G. Recolor X, Y, Z, P and G

Case 3• X’s Parent is Red (so Grandparent is Black)

and X and P are opposite children– Rotate P around G– Color P black– Color G red

• Again note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

Case 3 Diagrams (1 of 2)

X

ZY

P

G

U

S

X

YS

P

G

U

Z

Step 1 – recolor X, Y and Z. Rotate X around P.

Case 3 Diagrams (2 of 2)

X

YS

P

G

U

Z

P

YS

X

G

UZ

Step 2 – Rotate X around G. Recolor X and G

An exercise – insert F

D

T

W

ZV

L

J P

E K

Top-Down Insert Summary

P

X

Y Z

Case 1

P is Black

Just RecolorP

X

Y Z

Case 2P is RedX & P both left/right

P

X

Y Z

G

P

X

Y Z

GRecolorX,Y,Z

P

X

Y Z

GRotate P around GRecolor P,G

Case 3P is RedX and P are opposite children

P

X

Y Z

GRecolor X,Y,Z Rotate X around P

X

P

Y Z

G

Rotate X around GRecolor X, G

Recolor X,Y,Z

X

P

Y Z

G

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Insertion Practice

Insert the values 2, 1, 4, 5, 9, 3, 6, 7 into an initially empty Red-Black Tree

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Top-Down Insertion

An alternative to this “bottom-up” insertion is “top-down” insertion.

Top-down is iterative. It moves down the tree, “fixing” things as it goes.

What is the objective of top-down’s “fixes”?

Red Black Trees

Top-Down Deletion

Recall the rules for BST deletion

1. If vertex to be deleted is a leaf, just delete it.2. If vertex to be deleted has just one child,

replace it with that child3. If vertex to be deleted has two children,

replace the value of by it’s in-order predecessor’s value then delete the in-order predecessor (a recursive step)

What can go wrong?

1. If the delete node is red?

Not a problem – no RB properties violated

2. If the deleted node is black?

If the node is not the root, deleting it will change the black-height along some path

The goal of T-D Deletion

• To delete a red leaf

• How do we ensure that’s what happens?– As we traverse the tree looking for the leaf to

delete, we change every node we encounter to red.

– If this causes a violation of the RB properties, we fix it

Bottom-Up vs. Top-Down

• Bottom-Up is recursive– BST deletion going down the tree (winding up the

recursion)– Fixing the RB properties coming back up the tree

(unwinding the recursion)

• Top-Down is iterative– Restructure the tree on the way down so we don’t

have to go back up

Terminology

• Matching Weiss text section 12.2– X is the node being examined– T is X’s sibling– P is X’s (and T’s) parent– R is T’s right child– L is T’s left child

• This discussion assumes X is the left child of P. As usual, there are left-right symmetric cases.

Basic Strategy

• As we traverse the tree, we change every node we visit, X, to Red.

• When we change X to Red, we know– P is also Red (we just came from there)– T is black (since P is Red, it’s children are Black)

Step 1 – Examine the root

1. If both of the root’s children are Blacka. Make the root Redb. Move X to the appropriate child of the rootc. Proceed to step 2

2. Otherwise designate the root as X and proceed to step 2B.

Step 2 – the main case

As we traverse down the tree, we continually encounter this situation until we reach the node to be deleted

X is Black, P is Red, T is Black

We are going to color X Red, then recolor other nodes and possibly do rotation(s) based on the color of X’s and T’s children

2A. X has 2 Black children2B. X has at least one Red child

P

TX

Case 2AX has two Black Children

2A1. T has 2 Black Children

2A2. T’s left child is Red

2A3. T’s right child is Red

** if both of T’s children are Red, we can do either 2A2 or 2A3

Case 2A1X and T have 2 Black Children

P

TXP

TX

Just recolor X, P and T and move down the tree

Case 2A2

P

TX

L

X has 2 Black Children and T’s Left Child is Red

Rotate L around T, then L around PRecolor X and P then continue down the tree

L1 L2

P T

X

L

L1 L2

Case 2A3

P

TX

X has 2 Black Children and T’s Right Child is Red

Rotate T around PRecolor X, P, T and R then continue down the tree

R1 R2

P R

X

T

R2R1

R

L L

Case 2BX has at least one Red child

Continue down the tree to the next levelIf the new X is Red, continue down againIf the new X is Black (T is Red, P is Black)

Rotate T around PRecolor P and TBack to main case – step 2

Case 2B Diagram

P

X T

Move down the tree.

P

X T

P

T X

If move to Black child (2B2)Rotate T around P; Recolor P and TBack to step 2, the main case

If move to the Red child (2B1) Move down again

Step 3

Eventually, find the node to be deleted – a leaf or a node with one non-null child that is a leaf.

Delete the appropriate node as a Red leaf

Step 4Color the Root Black

Example 1Delete 10 from this RB Tree

15

17

16 20

23181310

7

12

6

3

Step 1 – Root has 2 Black children. Color Root Red

Descend the tree, moving X to 6

Example 1 (cont’d)

15

17

16 20

23181310

7

12

6

3

One of X’s children is Red (case 2B). Descend down the tree, arriving at 12. Since the new X (12) is also Red (2B1), continue down the tree, arriving at 10.

X

Example 1 (cont’d)

15

17

16 20

23181310

7

12

6

3

Step 3 -Since 10 is the node to be deleted, replace it’s value with the value of it’s only child (7) and delete 7’s red node

X

Example 1 (cont’d)

15

17

16 20

2318137

12

6

3

The final tree after 7 has replaced 10 and 7’s red node deleted and (step 4) the root has been colored Black.

Example 2Delete 10 from this RB Tree

15

17

16 20

1310

12

6

3

42

Step 1 – the root does not have 2 Black children.

Color the root red, Set X = root and proceed to step 2

Example 2 (cont’d)

15

17

16 20

1310

12

6

3

42

X

X has at least one Red child (case 2B). Proceed down the tree, arriving at 6. Since 6 is also Red (case 2B1), continue down the tree, arriving at 12.

Example 2 (cont’d)

15

17

16 20

1310

12

6

3

42

X

X has 2 Black children. X’s sibling (3) also has 2 black children.Case 2A1– recolor X, P, and T and continue down the tree, arriving at 10.

P

T

Example 2 (cont’d)

15

17

16 20

1310

12

6

3

42

P

X T

X is now the leaf to be deleted, but it’s Black, so back to step 2.X has 2 Black children and T has 2 Black children – case 2A1

Recolor X, P and T. Step 3 -- Now delete 10 as a red leaf.Step 4 -- Recolor the root black

Example 2 Solution

15

17

16 20

13

12

6

3

42

Example 3Delete 11 from this RB Tree

15

1311

12

10

5

73

6 9 2 4

Valid and unaffected Right subtree

Step 1 – root has 2 Black children. Color Root red.

Set X to appropriate child of root (10)

Example 3 (cont’d)

15

1311

12

10

5

73

6 9 2 4

X

X has one Red child (case 2B)

Traverse down the tree, arriving at 12.

Example 3 (cont’d)

15

1311

12

10

5

73

6 9 4

X

Since we arrived at a black node (case 2B2) assuring T is red and P is black), rotate T around P, recolor T and P

Back to step 2

P

T

2

Example 3 (cont’d)

15

1311

12 10

5

73

6 9 4

X

P

T

2

Now X is Black with Red parent and Black sibling.X and T both have 2 Black children (case 2A1)Just recolor X, P and T and continue traversal

Example 3 (cont’d)

15

1311

12 10

5

73

6 9 4X

P

T 2

Having traversed down the tree, we arrive at 11, the leaf to be deleted, but it’s Black, so back to step 2.X and T both have two Black children. Recolor X, P and T.Step 3 -- delete 11 as a red leaf. Step 4 -- Recolor the root black

Example 3 Solution

13

12 10

5

73

6 9 4 2

15