Introduction to
Algorithms
Jiafen Liu
Sept. 2013
Today’s Tasks
Balanced Search Trees
• Red-black trees
– Height of a red-black tree
– Rotations
– Insertion
Balanced Search Trees
• Balanced search tree: A search-tree
data structure for which a height of O(lgn)
is guaranteed when implementing a
dynamic set of n items.
• Examples:
– AVL Tree
– 2-3-4 Tree
– B Tree
– Red-black Tree
Red-black trees
• This data structure requires an extra 1-bit
color field in each node.
• Red-black properties:
– Every node is either red or black.
– The root and leaves (NIL’s) are black.
– If a node is red, then its parent is black.
– All simple paths from any node x to a
descendant leaf have the same number of
black nodes = black-height(x).
Example of a red-black tree
• Convention: black-height of x does not
include x itself.
Example of a red-black tree
• It could have a bunch of blacks, but it will
never repeat two reds in a row.
Goals of red-black trees
• there are a couple of goals that we are
trying to achieve.
– These properties should force the tree to
have logarithmic height, O(lgn) height.
– The other desire we have from these
properties is that they are easy to maintain.
• We can create a tree in the beginning that has this
property.
• A perfectly balanced binary tree with all nodes
black will satisfy those properties.
• The tricky part is to maintain them when we make
changes to the tree.
Height of a red black tree
• Let's look at the height of a red black tree.
And we will start to see where these
properties come from.
• Theorem. A red-black tree with n keys
has height h ≤ 2 lg(n + 1) .
– Can be proved by induction, as in the book
P164.
– Another informal way: intuition.
Intuition
• Merge red nodes into their black parents.
• This process produces another type of
balanced search tree: 2-3-4 tree.
– Any guesses why it's called a 2-3-4 tree?
– Another nice property:
• All of the leaves have the same depth. Why?
• By Property 4
height of merged tree h'
• Now we will prove the height of merged
tree h' .
• The first question is how many leaves are
there in a red-black tree?
– #(internal nodes) +1
– It can also be proved by induction. Try it.
Height of Red-Black Tree
•
•
•
•
•
•
The number of leaves in each tree is n + 1
2h' ≤ n + 1 ≤ 4h'
h' ≤ lg(n + 1)
How to connect h' and h?
We also have h' ≥ 1/2 h
h ≤ 2 lg(n + 1).
Query operations
• Corollary. The queries SEARCH, MIN,
MAX, SUCCESSOR, and PREDECESSOR all
run in O(lgn) time on a red-black tree with
n nodes.
Modifying operations
• The operations INSERT and DELETE
cause modifications to the red-black tree.
• How to INSERT and DELETE a node in
Red Black Tree?
– The first thing we do is just use the BST
operation.
– They will preserve the binary search tree
property, but don't necessarily preserve
balance.
Modifying operations
• The operations INSERT and DELETE
cause modifications to the red-black tree.
• How to INSERT and DELETE a node in
Red Black Tree?
– The second thing is to set color of new
internal node, to preserve property 1.
– We can color it red, and property 3 does not
hold.
– The good news is that property 4 is still true.
How to fix property 3?
• We are going to move the violation of
property 3 up the tree.
• IDEA: 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.
• we have to restructure the links of the tree
via “rotation”.
Rotations
• Rotations maintain the inorder ordering of
keys.
a ∈α, b ∈β, c ∈γ ⇒ a ≤ A ≤ b ≤ B ≤ c.
• A rotation can be performed in O(1) time.
– Because we only change a constant number
of pointers.
Implement of Rotation
• P166
Example
• Insert 15
Example
• Insert 15
• Color it red
Example
• Insert 15
• Color it red
• Handle Property 3 by recoloring
Example
• Insert 15
• Color it red
• Handle Property 3 by recoloring
Example
• Insert 15
• Color it red
• Handle Property 3 by recoloring
• Failed！
RIGHT-ROTATE
• RIGHT-ROTATE(18)
• It turns out to be?
RIGHT-ROTATE
• We are still in trouble between 10 and 18. But
made this straighter.
• It doesn't look more balanced than before.
• We can not resolve by recoloring. Then?
LEFT-ROTATE
• LEFT-ROTATE(7)
• It turns out to be?
LEFT-ROTATE
• LEFT-ROTATE(7) and recoloring at the
same time
• It satisfy all the 4 properties
Pseudocode of insertion
RB-INSERT(T, x)
TREE-INSERT(T, x)
color[x] ← RED // only RB property 3 can be violated
while x ≠ root[T] and color[x] = RED
do if p[x] = left[ p[ p[x] ]
then y ← right[ p[ p[x] ] // y = aunt or uncle of x
if color[y] = RED
then ⟨Case 1⟩
Case 1
• Let
• All
denote a subtree with a black root.
have the same black-height.
How?
• Push C’s black onto A and D, and recurse.
• Caution: We don't know whether B was the right
child or the left child. In fact, it doesn't matter.
Pseudocode of insertion
RB-INSERT(T, x)
TREE-INSERT(T, x)
color[x] ← RED // only RB property 3 can be violated
while x ≠ root[T] and color[x] = RED
do if p[x] = left[ p[ p[x] ]
then y ← right[ p[ p[x] ] // y = aunt/uncle of x
if color[y] = RED
then ⟨Case 1⟩
else if x = right[p[x]] //y is red, x,y have a zigzag
then ⟨Case 2⟩ // This falls into case 3
Case 2
How?
• Now we have a zigzig.
• we may want a straight path between x.
Pseudocode of insertion
RB-INSERT(T, x)
TREE-INSERT(T, x)
color[x] ← RED // only RB property 3 can be violated
while x ≠ root[T] and color[x] = RED
do if p[x] = left[ p[ p[x] ]
then y ← right[ p[ p[x] ] // y = aunt/uncle of x
if color[y] = RED
then ⟨Case 1⟩
else if x = right[p[x]]
then ⟨Case 2⟩ // This falls into case 3
⟨Case 3⟩ //straight path
Case 3
How?
• Done! No more violations of RB property 3
are possible.
• Property 4 is also preserved.
Pseudocode of insertion
RB-INSERT(T, x)
TREE-INSERT(T, x)
color[x] ← RED // only RB property 3 can be violated
while x ≠ root[T] and color[x] = RED
do if p[x] = left[ p[ p[x] ]
then y ← right[ p[ p[x] ] // y = aunt/uncle of x
if color[y] = RED
then ⟨Case 1⟩
else if x = right[p[x]]
then ⟨Case 2⟩ // This falls into case 3
⟨Case 3⟩
else ⟨“then” clause with “left” and “right” swapped⟩
color[root[T]] ← BLACK
Analysis
• Go up the tree performing Case 1, which
only recolors nodes.
• If Case 2 or Case 3 occurs, perform 1 or 2
rotations, and terminate.
• Running time of insertion?
– O(lgn) with O(1) rotations.
• RB-DELETE — same asymptotic running
time and number of rotations as RBINSERT (see textbook).