Transcript Lecture 24

Algorithms
CSCI 235, Fall 2015
Lecture 24
Red Black Trees
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Properties of Red-black trees
1.
2.
3.
4.
5.
Every node is either red or black
The root is black
Every leaf (nil(T)) is black
If a node is red, both children are black
For each node, all paths from that node to the descendent leaves
contain the same number of black nodes.
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NIL
NIL
Usually omitted
when drawing
trees
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NIL
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NIL
NIL
NIL
NIL
2
Red-Black Insertion
Insertion in a Red-Black Tree starts the same way as insertion into
a regular binary-search-tree.
Insert node z at the bottom of the tree by following the branches
from root to leaf in the correct order.
Then:
1) Color node z red.
2) Call RB-Insert-Fixup(T, z) to regain Red-Black Tree
properties.
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RB-Insert(T, z) pseudocode
RB-Insert(T, z)
...
{Insert z into T, just like Binary-Search-Tree}
...
left[z] <- nil[T]
right[z] <- nil[T]
color[z] <- RED
RB-Insert-Fixup(T, z)
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Fixing up the tree
What might need fixing?
Property
Description
1
Every node in the tree is red or black
2
The root is black
3
Every leaf (nil[T]) is black
4
If a node is red, both children are black
5
All paths from any given node to the
descendant leaves contain the same number
of black nodes.
True?
yes
?
yes
?
yes
Properties, 1, 3 and 5 still hold.
Therefore, we only need to check properties 2 and 4 and fix them if
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they are violated.
Things we know
The following are true at the start of RB-Insert-Fixup:
a) Node z is red.
b) if p[z] is the root, then p[z] is black. Therefore, if p[z] is red,
then p[p[z]] exists.
c) If property 4) is violated, then it is because p[z] is red.
d) If property 2) is violated, it is because z is the root of the tree.
As we fix up the tree, we keep these facts the same after each iteration
of the fixup (i.e. they are loop invariant).
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Cases to Consider
If property 2 is violated, then z is the root. Color z black and we
are done.
Cases to consider for Property 4:
Case 1: z's parent's sibling (i.e. z's uncle) is also red.
Case 2: z's uncle is black and z is a right child.
Case 3: z's uncle is black and z is a left child.
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Case 1
Case 1: z's uncle is also red.
C z
C
uncle
D y
A
a
B z
b
g
A
a
D
B
b
g
a) Color both p[z] and uncle of z black.
b) Color p[p[z]] red (to maintain equal numbers of black nodes on
all paths.
c) Move z pointer up two levels (to p[p[z]]).
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d) Repeat check for violation of properties 2 or 4.
Pseudocode for case 1
while color[p[z]] = RED do
{violation of property 4}
if p[z] = left[p[p[z]]] then
y <- right[p[p[z]]]
{y is uncle}
if color[y] = RED then
{case 1 applies}
color[p[z]] <- BLACK
color[y] <- BLACK
color[p[p[z]]] <- RED
z <- p[p[z]]
...{consider cases 2 and 3}
else
{else clause is the same except left and right exchanged}
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Case 2
Case 2: z's uncle is black and z is a right child.
Use left rotation to convert to case 3.
Case 3: z's uncle is black and z is a left child.
C
uncle
D y
A
a
C
B z
B
g
A z
Case 2
b
g
a
if z = right[p[z]] then
z <- p[z]
Left-Rotate(T, z)
{Proceed with case 3}
D
uncle
y
Case 3
b
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Case 3
Case 3: z's uncle is black and z is a left child.
B
C
uncle
D y
B
A z
a
g
zA
a
Case 3
C
b
g
D
b
1) Color p[z] black.
2) Color p[p[z]] red.
3) Perform right rotation.
After fixing case 3, we are done. There are no more violations (Why?)
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Pseudocode for case 3
color[p[z]] <- BLACK
color[p[p[z]]] <- RED
Right-Rotate(T, p[p[z]])
{End of While loop}
{After the end of the while loop, set the root color to black:}
color[root[T]] <- BLACK {fix any violations of property 2}
Running time of RB-Insert-Fixup = ?
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