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CS 361, Lecture 17
Jared Saia
University of New Mexico
Outline
Binary Search Trees
Red-Black Trees
Class Evaluation
Hash Tables
Hash Tables implement the Dictionary ADT, namely:
Insert(x) -
O
(1) expected time, Θ(
n ) worst case
Lookup(x) -
O
(1) expected time, Θ(
n ) worst case
Delete(x) -
O
(1) expected time, Θ(
n ) worst case
Red-Black Trees
tionary ADT, namely: Red-Black trees (a kind of binary tree) also implement the Dic-
Insert(x) -
O
(log
(^) n
) time
Lookup(x) -
O
(log
(^) n
) time
Delete(x) -
O
(log
(^) n
) time
Why BST?
Q: When would you use a Search Tree?
(e.g. “mission critical” code)A1: When need a hard guarantee on the worst case run times
(e.g.A2: When want something more dynamic than a hash table
don’t want to have to enlarge a hash table when the
load factor gets too large)
A3:
Search trees can implement some other important op-
erations...
Search Tree Operations
Insert
Lookup
Delete
Minimum/Maximum
Predecessor/Successor
What is a BST?
It’s a binary tree
and right childrenEach node holds a key and record field, and a pointer to left
Binary Search Tree Property
is maintained
Binary Search Tree Property
Let
x
be a node in a binary search tree. If
y
is a node in the
left subtree of
x , then key(y)
key(x).
If
y
is a node in the
right subtree of
x
then key(x)
key(y)
Analysis
Correctness?
Run time?
Search in BT
Tree-Search(x,k){
if (x=nil) or (k = key(x)){
return x;
if (k
Tree Min/Max
Tree
Minimum(x):
Return
the
leftmost
child
in
the
tree
rooted at x
Tree Maximum(x):
Return the rightmost child in the tree
rooted at x
Tree-Successor
Tree-Successor(x){
if (right(x) != null){
return Tree-Minimum(right(x));
while (y!=null and x!=left(y)){y = parent(x);}
y = parent(y);x = y;
return y;}
Successor Intuition
Case 1:
If right subtree of
x
is non-empty, successor(x) is
just the leftmost node in the right subtree
Case 2:
If the right subtree of
x
is empty and
x
has a suc-
cessor, then successor(x) is the lowest ancestor of
x
whose
left child is also an ancestor of
x .
Insertion
Insert(T,x)
- Let
r
be the root of
T
- Do Tree-Search(r,key(x)) and let
p
be the last node pro-
cessed in that search
- If
p
is nil (there is no tree), make
x
the root of a new tree
- Else if key(x)
p, make
x
the left child of
p , else make
x
the right child of
p
Analysis
“Shut up brain or I’ll poke you with a Q-Tip” - Homer Simpson
Note that the average depth of a node in
T
is
n 1
x ∑ ∈ T
d ( x, T
n 1 P (^) ( T (^) )
Thus we want to show that
P
T
O
n (^) log
(^) n
)
Analysis
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 (^) −
(^) n
(^) −
P Analysis
n )
n 1
n − 1
i=0 ∑
P
i ) +
P
n (^) −
(^) i (^) −
(^) n
(^) −
(^) 1)
n 1 ( n − 1
i=0 ∑
P
i ) +
P
n (^) −
(^) i (^) −
n 1 ( n − 1
i=0 ∑
n (^) −
n 1 ( n − 1
i=0 ∑
P
i ) +
P
n (^) −
(^) i (^) −
n )
n 2 ( n − 1
∑
k
P
k )) + Θ(
n )
Analysis
We have
P
n ) =
n 2 ( ∑
n − 1
k
P
k )) + Θ(
n )
This is the same recurrence for randomized Quicksort
this recurrence isIn your hw (problem 7-2), you showed that the solution to
P
n ) =
O
n (^) log
(^) n
)
Take Away
P
n ) is the expected total depth of all nodes in a randomly
built binary tree with
n
nodes.
We’ve shown that
P
n ) =
O
n (^) log
(^) n
)
There are
n
nodes total
Thus the expected average depth of a node is
O
(log
(^) n
)
Take Away
binary tree isThe expected average depth of a node in a randomly built
O
(log
(^) n
)
expected timeThis implies that operations like search, insert, delete take
O
(log
(^) n
) for a randomly built binary tree
Warning!
search treeIn many cases, data is not inserted randomly into a binary
I.e. many binary search trees are not “randomly built”
tree in almost sorted orderFor example, data might be inserted into the binary search
Then
the
BST
would
not
be
randomly
built,
and
so
the
expected average depth of the nodes would not be
O
(log
(^) n
)
