1. CS 16: Balanced Trees CS 16: Balanced Trees
2-3-4 Trees Revealed
2-3-4 Trees and Red-
Black Trees • Nodes store 1, 2, or 3 keys and have
2, 3, or 4 children, respectively
• All leaves have the same depth
k
beh nr
a cd fg i lm p sx
1
-- log ( N + 1 ) ≤ height ≤ log ( N + 1 )
-
2
erm 204 erm 205
CS 16: Balanced Trees CS 16: Balanced Trees
2-3-4 Tree Nodes Why 2-3-4?
• Introduction of nodes with more than 1 key, • Why not minimize height by maximizing
and more than 2 children children in a “d-tree”?
• Let each node have d children so that we
a
2-node: get O(log d N) search time! Right?
logdN = log N/log d
• same as a binary node
<a >a
a b
3-node:
• 2 keys, 3 links <a >b
>a
<b • That means if d = N1/2, we get a height of 2
• However, searching out the correct child
a b c on each level requires O(log N1/2) by
4-node: binary search
• 3 keys, 4 links
• 2 log N1/2 = O(log N) which is not as good
<a >c as we had hoped for!
>a >b • 2-3-4-trees will guarantee O(log N) height
<b <c using only 2, 3, or 4 children per node
erm 206 erm 207

2. CS 16: Balanced Trees CS 16: Balanced Trees
Insertion into 2-3-4 Trees Top Down Insertion
• In our way down the tree, whenever we
• Insert the new key at the lowest internal reach a 4-node, we break it up into two 2-
node reached in the search nodes, and move the middle element up
into the parent node
e e
• 2-node becomes 3-node n fn
g d dg
dfg g
d
• 3-node becomes 4-node
f dg d f g
• Now we can perform the
insertion using one of the fn
previous two cases
• Since we follow this
method from the root down de g
• What about a 4-node?
to the leaf, it is called
• We can’t insert another key! top down insertion
erm 208 erm 209
CS 16: Balanced Trees CS 16: Balanced Trees
Splitting the Tree
g n
As we travel down the tree, if we encounter any
4-node we will break it up into 2-nodes. This
guarantees that we will never have the problem c t
of inserting the middle element of a former 4-
node into its parent 4-node.
a fil pr x
g c n t
pr x Whoa, cowboy n
a fil
g c t
g n
Whoa, cowboy
a fil pr x
c t
a fil pr x
erm 210 erm 211

3. CS 16: Balanced Trees CS 16: Balanced Trees
n
Time Complexity of Insertion
g c t in 2-3-4 Trees
Whoa, cowboy
a fil pr x Time complexity:
• A search visits O(log N) nodes
n
• An insertion requires O(log N) node splits
• Each node split takes constant time
g ci t
• Hence, operations Search and Insert each
take time O(log N)
a f l pr x Notes:
• Instead of doing splits top-down, we can
n perform them bottom-up starting at the in-
sertion node, and only when needed. This
ci t is called bottom-up insertion.
• A deletion can be performed by fusing
nodes (inverse of splitting), and takes
a fg l pr x O(log N) time
erm 212 erm 213
CS 16: Balanced Trees CS 16: Balanced Trees
Beyond 2-3-4 Trees Red-Black Tree
A red-black tree is a binary search tree with the
What do we know about 2-3-4 Trees? following properties:
• edges are colored red or black
• Balanced Siskel • no two consecutive red edges
on any root-leaf path
• O(log N) search time • same number of black edges
Ebert on any root-leaf path
(= black height of the tree)
• Different node structures Roberto • edges connecting leaves are black
Can we get 2-3-4 tree advantages in a binary Black Red
Edge Edge
tree format???
Welcome to the world of Red-Black Trees!!!
erm 214 erm 215

4. CS 16: Balanced Trees CS 16: Balanced Trees
2-3-4 Tree Evolution More Red-Black Tree
Properties
Note how 2-3-4 trees relate to red-black trees N # of internal nodes
L # leaves (= N + 1)
2-3-4 H height
Red-Black
B black height
Property 1: 2B ≤ N + 1 ≤ 4B
or
1
Property 2: -- log ( N + 1 ) ≤ B ≤ log ( N + 1 )
-
2
Property 3: log ( N + 1 ) ≤ H ≤ 2 log ( N + 1 )
Now we see red-black trees are just a way of
representing 2-3-4 trees! This implies that searches take time O(logN)!
erm 216 erm 217
CS 16: Balanced Trees CS 16: Balanced Trees
Insertion into Red-Black Trees Insertion - Plain and Simple
1.Perform a standard search to find the leaf
where the key should be added
Let:
2.Replace the leaf with an internal node with n be the new node
the new key p be its parent
3.Color the incoming edge of the new node g be its grandparent
red
4.Add two new leaves, and color their
Case 1: Incoming edge of p is black
incoming edges black
5.If the parent had an incoming red edge, we
now have two consecutive red edges! We No violation
g
must reorganize tree to remove that
violation. What must be done depends on
the sibling of the parent. p STOP!
R R n
Pretty easy, huh?
G G
Well... it gets messier...
erm 218 erm 219

5. CS 16: Balanced Trees CS 16: Balanced Trees
Restructuring More Rotations
Case 2: Incoming edge of p is red, and
its sibling is black Similar to a right rotation, we can do a
“left rotation”...
Uh oh!
g
Much g
Better! p
p p
p
n g
n g n
n
We call this a “right rotation”
• No further work on tree is necessary
Simple, huh?
• Inorder remains unchanged
• Tree becomes more balanced
• No two consecutive red edges!
erm 220 erm 221
CS 16: Balanced Trees CS 16: Balanced Trees
Double Rotations Last of the Rotations
What if the new node is between its parent and And this would be called a
grandparent in the inorder sequence? “right-left double rotation”
We must perform a “double rotation”
(which is no more difficult than a “single” one)
g n g n
p p g p g p
n n
This would be called a
“left-right double rotation”
erm 222 erm 223

6. CS 16: Balanced Trees CS 16: Balanced Trees
Bottom-Up Rebalancing Summary of Insertion
Case 3: Incoming edge of p is red and its
sibling is also red
• If two red edges are present, we do either
• a restructuring (with a simple or double
g g rotation) and stop, or
• a promotion and continue
p p • A restructuring takes constant time and is
performed at most once. It reorganizes an
off-balanced section of the tree.
n n
• Promotions may continue up the tree and
are executed O(log N) times.
• We call this a “promotion” • The time complexity of an insertion is
• Note how the black depth remains un- O(logN).
changed for all of the descendants of g
• This process will continue upward beyond
g if necessary: rename g as n and repeat.
erm 224 erm 225
CS 16: Balanced Trees CS 16: Balanced Trees
An Example A Cool Example
Start by inserting “REDSOX” into an empty tree C E
D R
E
O S
D R
X
O S
E
X
D R
Now, let’s insert “C U B S”...
C O S
X
erm 226 erm 227

7. CS 16: Balanced Trees CS 16: Balanced Trees
E
An Unbelievable Example
U E D R
D R
C O S
C O S
Double Rotation X
E X
U
D R E
C O S D R
X C O U
Oh No!
What should we do? S X
U
erm 228 erm 229
CS 16: Balanced Trees CS 16: Balanced Trees
A Beautiful Example E
E
B D R
D R
Rotation C O U
C O U
B S X
S X
E E
D R C R
What
now? C O U B D O U
B S X S X
erm 230 erm 231

8. CS 16: Balanced Trees CS 16: Balanced Trees
E
A Super Example
E
S C R
C R
B D O U
Use the
B D O U Bat-Promoter!!
S X
S X
E E S
C R C R BIFF!
B D O U B D O U
We could’ve
Holy Consecutive placed it on
S
Red Edges, Batman! S X X
either side
S S
erm 232 erm 233
CS 16: Balanced Trees
E
C R Rotation
B D O U
S X
S
R
The SUN lab
and Red-Bat
trees are safe... E U
...for now!!!
C O S X
B D S
erm 234