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CS 361, Lecture 20
Jared Saia
University of New Mexico
Outline
Red-Black Trees
Randomly Built BSTs
depth of a node in a randomly built tree?”We want to answer the question: “What will be the average
Define the random variable
X
to be the depth of a node
chosen uniformly at random in the tree
X
takes on
n
possible values, it takes on each value with
probability 1
/n
Our Problem
For a tree
T
and node
x , let
d ( x, T
(^) ) be the depth of node
x
in
T
Define the total path length,
P
T
(^) ), to be the sum over all
nodes
x
in
T
of
d ( x, T
Then
E
X
n 1
x ∑ ∈ T
d ( x, T
n 1 P (^) ( T (^) )
Thus we want to show that
P
T
O
n (^) log
(^) n
)
Analysis
“Shut up brain or I’ll poke you with a Q-Tip” - Homer Simpson
Let
T
l ,
T r
be the left and right subtrees of
T
respectively.
Let
n
be the number of nodes in
T
Then
P
T
P
T
l ) +
P
T
r ) +
(^) n
(^) 1. Why?
Analysis
Let
P
n ) be the expected total depth of all nodes in a ran-
domly built binary tree with
n
nodes
Note that for all
i , 0
i ≤
n (^) −
(^) 1, the probability that
T
l has
i
nodes and
T
r
has
n (^) −
i (^) −
(^) 1 nodes is 1
/n
Thus
P
n ) =
n 1 ∑
n − 1
i
P
i ) +
P
n (^) −
i −
(^) 1) +
(^) n
(^) −
P Analysis
n )
n 1
n − 1
i=0 ∑
P
i ) +
P
n (^) −
i (^) −
(^) n
(^) −
n 1 ( n − 1
i=0 ∑
P
i ) +
P
n (^) −
i −
(^) 1)) +
n 1 ( n − 1
i=0 ∑
n (^) −
n 1 ( n − 1
i=0 ∑
P
i ) +
P
n (^) −
i −
(^) 1)) + Θ(
n )
n 2 ( n − 1
k=1 ∑
P
k )) + Θ(
n )
Analysis
We have
P
n ) =
n 2 ( ∑
n − 1
k
P
k )) + Θ(
n )
This is the same as the recurrence for randomized Quicksort
rence isRecall from hw problem 7-2, that the solution to this recur-
P
n ) =
O
n (^) log
(^) n
)
What is a RB-Tree
A RB-Tree is a balanced binary search tree
The
height
of
the
tree
is
always
O
(log
(^) n
)
where
n
is
the
number of nodes in the tree
RB Trees
and parent pointerEach node has a “color” field in addition to a key, left, right,
sponding pointer field will contain the value NILIf the child or parent of a node does not exist, the corre-
internal nodes of the tree(leaves) of the tree, and say that all key-bearing nodes areWe will say that these NIL’s are pointers to external nodes
Red-Black Properties
A BST is a red-black tree if it satisfies the RB-Properties
- For each node, all paths from the node to descendant leaves4. If a node is red, then both its children are black3. Every leaf (NIL) is black2. The root is black1. Every node is either red or black
contain the same number of black nodes
Example RB-Tree
Black Height
Black-height
of a node
x , bh(x) is the number of black nodes
on any path from, but not including
x
down to a leaf node.
paths have the same number of black nodesNote that the black-height of a node is well-defined since all
the rootThe black-height of an RB-Tree is just the black-height of
Key Lemma
Lemma: A RB-Tree with
n
internal nodes has height at most
2 log(
n
- The subtree rooted at the nodeProof Sketch:
x
contains at least
bh
( x ) −
(^) 1 internal nodes
- For the root
r ,
bh
( r )
≥
h/
2, thus
n
h/
2
−
Taking
logs of both sides, we get that
h
≤
(^) 2 log(
n (^) + 1)
Proof
- The subtree rooted at the node
x
contains at least 2
bh
( x ) −
internal nodes. Show by induction on the height of
x .
BC: If the height of
x
is 0,
then
x
is a leaf,
and subtree
rooted at
x
does indeed contain 2
0
−
(^) 1 = 0 internal nodes
IH: For all nodes
y
of height less than
x , the subtree rooted
at
y
contains at least 2
bh
( y ) −
(^) 1 internal nodes.
IS: Consider a node
x
which is an internal node with two
children(all internal nodes have two children).
Each child
has black-height of either
bh
x ) or
bh
x ) −
1 (the former if
it is red, the latter if it is black).
Since the height of these
children is less than
x , we can apply the inductive hypothesis
to conclude that each child has at least 2
bh
( x ) − 1
−
(^) 1 internal
nodes. This implies that the subtree rooted at
x
has at least
bh
( x ) − 1 −^
(^) 1) + (
bh
( x ) − 1 −^
(^) 1) + 1 = 2
bh
( x ) −^
(^) 1 internal nodes.
This proves the claim.
Maintenance?
tained?How do we ensure that the Red-Black Properties are main-
holds?we re-arrange the new tree so that the Red-Black PropertyI.e. when we insert a new node, what do we color it? How do
How about for deletions?
In-Class Exercise
words, Show that Left-Rotate(x) maintains the BST Property. In other
show
that
if
the
BST
Property
was true
for
the
tree
after the operation.before the Left-Rotate(x) operation, then it’s true for the tree
Show that after rotation,
the BST property holds for the
entire subtree rooted at
x
Show that after rotation,
the BST property holds for the
subtree rooted at
y
the entire treeNow argue that after rotation, the BST property holds for
