Lecture 2 Red-Black Trees. 8/3/2007 UMBC CSMC 341 Red-Black-Trees-1 2 Red-Black Trees Definition: A...

Lecture 2

Red-Black Trees

8/3/2007UMBC CSMC 341 Red-Black-

Trees-1 2

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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Theorem :– In a red-black tree, at least half the nodes on any path from the root to a NULL must be Black.

Proof – If there is a Red node on the path, there must be a corresponding Black node.

Algebraically this theorem means

bh( x ) ≥ h/2

Claim

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Claim (cont’d) example

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17 41

30 47

38 50

NIL NIL

NIL

NIL NIL NIL NIL

NIL

h = 4bh = 2

h = 3bh = 2

h = 2bh = 1

h = 1bh = 1

h = 1bh = 1

h = 2bh = 1 h = 1

bh = 1

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Rotations Operations for re-structuring the tree after insert

and delete operations on red-black trees

Rotations take a red-black-tree and a node within

the tree and: Together with some node re-coloring they help restore the

red-black-tree property

Change some of the pointer structure

Do not change the binary-search tree property

Two types of rotations: Left & right rotations

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Left Rotations Assumptions for a left rotation on a node x:

The right child of x (y) is not NIL

Idea: Pivots around the link from x to y Makes y the new root of the subtree x becomes y’s left child y’s left child becomes x’s right child

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Example: LEFT-ROTATE

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LEFT-ROTATE(T, x)1. y ← right[x] ►Set y

2. right[x] ← left[y] ► y’s left subtree becomes x’s right subtree

3. if left[y] NIL4. then p[left[y]] ← x ► Set the parent relation from left[y] to x

5. p[y] ← p[x] ► The parent of x becomes the parent of y

6. if p[x] = NIL7. then root[T] ← y8. else if x = left[p[x]]9. then left[p[x]] ← y10. else right[p[x]] ← y11. left[y] ← x ► Put x on y’s left

12. p[x] ← y ► y becomes x’s parent

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Right Rotations Assumptions for a right rotation on a node x:

The left child of y (x) is not NIL

Idea: Pivots around the link from y to x Makes x the new root of the subtree y becomes x’s right child x’s right child becomes y’s left child

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Example: RIGHT-ROTATE

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Insertion

Goal:

Insert a new node z into a red-black-tree

Idea:

Insert node z into the tree as for an ordinary

binary search tree

Color the node red

Restore the red-black-tree properties


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Insertion

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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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

Both parent and uncle are Red color parent and uncle Black, color grandparent Red. Point X to grandparent and check new situation.


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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

(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


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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

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

Insertion into a redblack tree


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Example: Insert x in tree. Color x red. Only red-black property 3 might be violated. Move the violation up the tree by recoloring until it can be fixed with rotations and recoloring.

Insert x =15.



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2

1 7

5 8

Insert 4 into this R-B Tree

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Example

11Insert 4

2 14

1 157

85

4

y

11

2 14

1 157

85

4

z

Case 1

y

z and p[z] are both redz’s uncle y is redz

z and p[z] are both redz’s uncle y is blackz is a right child

Case 2

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2

14

1

15

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8

5

4

z

yCase 3

z and p[z] are redz’s uncle y is blackz is a left child

112

141

15

7

85

4

z

Deletion

Recall the rules for BST deletion1. 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

Terminology

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.

Step 1 – Examine the root

1. If both of the root’s children are Blacka. Make the root Red

b. Move X to the appropriate child of the root

c. 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

TX

P

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 level

If the new X is Red, continue down again

If the new X is Black (T is Red, P is Black)

Rotate T around P

Recolor P and T

Back 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

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1620

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)

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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


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23181310

7

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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


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1620

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.


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1310

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3

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

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


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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.


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1620

1310

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3

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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


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1310

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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

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1620

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6

3

42


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6

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Valid and unaffected Right subtree

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

Set X to appropriate child of root (10)


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1311

12

10

5 73

6

9

2

4

X

X has one Red child (case 2B)

Traverse down the tree, arriving at 12.


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10

5

7

3

6

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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


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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


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1311

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5

7

3

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X

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

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Lecture 2 Red-Black Trees. 8/3/2007 UMBC CSMC 341 Red-Black-Trees-1 2 Red-Black Trees Definition: A...

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