Lecture-17-CS210-2012 (1).pptx

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Data Structures and Algorithms

(CS210/ESO207/ESO211)

Lecture 17

Red-Black trees Handling deletion of nodes

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RedBlackTree

Red-Black tree is a fullbinary search tree with each leaf as a nullnode and

satisfying the following properties at each stage.

Each node is colored redor black.

Each leaf is colored black and so is the root.

Every red node will have both its children black.

The number of blacknodes on a path from root to a leaf node is same for

every leaf. This parameter is called blackheight of the tree.

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A red-blacktree

3

root

2

28

46

67

25

5

31 41

35 49

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Handling Insertion in a Red BlackTree

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Aquick recap from the previous

lecture

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Rotation around a nodeAn important tool for balancing trees

5

v

u

xRight rotation Left rotation

v

u

x

p

Note that the tree T continues to

remain a BST even after rotation

around any node.

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Insertion in a red-black tree

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T

2

28

46

67

25

5

31 41

35 49 83

54

Insert(T,44) :Inserting 44into T.

44

Color imbalance

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Key pointsabout Insertion in a Red-Blacktree

Black Height of the tree is kept intactalways.

An insertion can potentially lead to color imbalance.

A color imbalance is handled using one or more of the following

tools:

1. Swappingof colors

2. Shiftingthe imbalance upward3. Rotationaround a node

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Handling Deletion in a Red BlackTree

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Notations to be used

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a blacknode

a rednode

a node whose color is either redor black

a BSTCould potentially be

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How to maintain a BST under deletion of

nodes ?

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Deletion in a BST isslightly harder than Insertion

(even ignoring the height balance factor)

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T

2

28

46

67

25

5

31 41

35 49 83

54

44

How to delete 28 ?

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Is deletion of a node easier for some cases ?

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T

2

28

46

67

25

5

31 41

35 49 83

54

44

Deletion of 31 is easy.

Can you see ?Deletion of 49 is also

easy.

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An insight

It is easier to maintain a BST under deletion if the node to be deleted has at

most one child which is non-leaf.

13

p

q

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An insight



14

q

p

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An insight



15

q

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An important question



Question: Can we transform every other case to the above case ?

Answer: ??

16

p

non-leafnon-leaf

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How to delete a node whose both children are non-leaves?

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T

2

28

46

67

25

5

31 41

35 49 83

54

44

How to delete 28 ?

Swap28 and25

and then delete 28

How to delete 46 ?

Swap46 and44

and then delete 46

Can you figure

out the strategy

now ?

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An important observation



Question: Can we transform every other case to the above case ?

Answer: by swapping value(p) with its predecessor.

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p

non-leafnon-leaf

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We need to handle deletion only for the

following case

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p

q

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How to maintain a red-blacktree under

deletion ?

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We shall first perform deletion like in a BST and

then restore all properties of red-black tree.

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Easy cases and difficult case

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p

q

Easy case Difficult case

p

q

p

q

Easy case:Change color of qto

black

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Handling the difficult case

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p

q

s

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23

q

sp

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q

s

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q s

r

As some students had noticed after the class that the subtree rooted at q

will actually be just a leaf node in the beginning. But we are not showing

it explicitly here. This is because we are depicting the most general case.

During the algorithm, we might shift the height imbalance upwards and

in that case the subtree rooted at q might not be a leaf node.

Notice that the number of

black nodes to each leaf

node in subtree(q) has

become oneless than any

other leaf node. We need

an algorithm to remove this

black-height imbalance.

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Handling the difficult case: An overview

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sisred sisblackreduction

one child of sis redBoth children of sare black

right(s)isredleft(s)isred and

right(s) isblackreduction

Eventually we need

to handle these two

cases only

Color ofs ?

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Reduction of

case s is red to case s is black

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Reduction of case s is red to case s is black

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q s

21

r

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q s

21

r

Left rotation

r

q

2

1

s

The new sibling of q is now a black

node. But the number of black nodes

to leaves of tree 2 have reduced by

one. What to do ?

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q s

21

r

Left rotation

r

q

2

1

s

Convince yourself that the number of

black nodes to any leaf of subtree(q)

or subtrees 1 and 2 is now the same

as before the rotation. And now the

sibling of q is black. So we are done.

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We just need to handle the case

s is black

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Handling the case when sis black

Case 1:both children of s are black

Case 2:at least one child of s is red

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q s

r

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Handling the case:

s is black and both children of s are black

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Handling the case:


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What if we swap colors of

s and r

q s

21

r

When r is red

q s

21

r

When r is black

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Handling the case:


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q s

21

r

q s

21

r

What if we change color

of s to red

When r is red When r is black

Swapping the colors leaves the number

of black nodes to the leaves of trees 1

and 2 unchanged. Interestingly, the

deficiency of one black node on the

path to the leaves of subtree rooted at

q is also compensated. So we may stop.

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Handling the case:


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q s

r

q s

r

When r is red When r is black

Changing color of sto redhas reduced

the number of black nodes on the path

to the root of subtree(q) by one. As a

result the imbalance of black height

haspropagatedto r. So we process

vertex r in a similar fashion.

Swapping the colors leaves the number

of black nodes to the leaves of trees 1

and 2 unchanged. Interestingly, the

deficiency of one black node on the

path to the leaves of subtree rooted at

q is also compensated. So we may stop.

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Handling the case:

s is black and one of its children is red

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There are two cases

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q s

r

When right(s) is red reduction

q s

r

When left(s) is red andright(s) isblack

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Handling the case: right(s) is red

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q s

21

r

Let color(r) be c

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q s

21

r

Left rotation

q

r

2

1

s

The number of black nodes on the path from root to

any leaf node of subtree(q) has increased by one

(this is good!), has remained unchanged for leaves of

tree 1, and is uncertain for leaves of tree 2(depends

upon c). How to get rid of this uncertainty ?

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q s

21

r

Left rotation

q

r

2

1

s

The number of black nodes on the path from

root to any leaf node 2 is now less by one

node. What to do ? (Hint: root of tree 2 is red)

Change color of root

of tree 2 to black and

we are done.

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q s

21

r

Left rotation

q

r

2

1

s

Convince yourself that left rotation at r,

followed by color swap of s and r, followed by

change of color of root of tree r removes the

imbalance of black height for all leaf nodes ofthe subtrees shown.

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Handling the caseleft(s) is redandright(s) is black

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Handling the case:

left(s) is red and right(s) is black

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q s

r

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Handling the case:


46

q s

3

1

r

2

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Handling the case:


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q s

3

1

r

2

Right rotation

q

s

3

1

r

2

The number of black nodes on the path from

root to any leaf node in tree 1 has now reduced

by one athough it is the same for trees 2 and 3.

What should we do ?

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Handling the case:


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q s

3

1

r

2

Right rotation

q

s

3

1

r

2

Notice that now the new sibling of qhas its right

child red. So we have effectively reduced the

current case to the case which we know how to

handle.

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Theorem: We can maintain red-black trees under deletion of

nodes in O(log n) time per insert/search operation where nis the

number of the nodes in the tree.

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Key pointsabout Deletion in a Red-Blacktree

3rdproperty of the red-black tree (no two consecutive red

nodes) is kept intactalways.

An insertion can potential lead to an imbalance of black

height(some leaf nodes having less number of black nodes on the pathfrom the root).

An imbalance of black height is handled using one or more of

the following tools:1. Swappingof colors

2. Shiftingthe imbalance upward

3. Rotationaround a node

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Points to ponder

As the nodes are being deleted, the black nodes will reduce in numbersand so will the black height. Among various cases we discussed for

handling deletion of nodes, figure out the case(s) which can potentially

lead to a reduction in the black height of the tree.

What is the worst case number of rotation operations during a singledeletion in a red black tree ?

What is the worst case number of rotation operations during a single

insertion in a red black tree ?

Lecture-17-CS210-2012 (1).pptx

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