Download Binary Search Tree

Introduction to
Algorithms
Jiafen Liu
Sept. 2013
Today’s Tasks
Binary Search Trees(BST)
• Analyzing height
• Informal Analysis
• Formal Proof
– Convexity lemma
– Jensen’s inequality
– Exponential height
Binary Search Tree
• Binary search tree is an popular data structure,
it supports several dynamic operations.
– Search
– Minimum
– Maximum
– Predecessor
– Successor
– Insert
– Delete
– Tree Walk
Data structure of binary search tree
• A BST tree can be represented by a
linked data structure in which each node
is an structure.
– Key field
– Satellite data
– Pointers: left, right, and parent
• For any node x, the keys in the left
subtree of x less than or equal to key[x],
and the keys in the right subtree of x are
bigger than key[x].
Different Shapes of BST
• 2 binary search trees contains the same 6 keys.
– (a) A binary search tree with height 3.
– (b) A binary search tree with height 5.
– Which one is better?
• Balanced Tree and Unbalanced Tree, What’s
the worst case of BST?
What’s the cost of basic operations?
• It takes Θ(n) time to walk an n-node
binary search tree.
• Other basic operations on a binary search
tree take time proportional to the height of
the tree Θ(h) .
Tree Walk
• If x is the root of an n-node subtree, then
the call INORDER-TREE-WALK(x) takes
Θ(n) time.
• We will prove by induction.
Proof of the theorem
Proof:
• Let T(n) denote the time taken this function.
• T(1) is a constant.
• For n > 1, suppose that function is called on a
node x whose left subtree has k nodes and
whose right subtree has n - k - 1 nodes.
T(n) = T(k) + T(n - k - 1) + d
• How to Solve it?
– substitution method
Search
TREE-SEARCH (x, k)
1 if x= NIL or k = key[x]
2 then return x
3 if k < key[x]
4 then return TREE-SEARCH(left[x], k)
5 else return TREE-SEARCH(right[x], k)
Minimum and maximum
TREE-MINIMUM (x)
1 while left[x] ≠ NIL
2
do x ← left[x]
3 return x
TREE-MAXIMUM(x)
1 while right[x] ≠ NIL
2
do x ← right[x]
3 return x
Successor and predecessor
TREE-SUCCESSOR(x)
1 if right[x] ≠ NIL
2
then return TREE-MINIMUM (right[x])
3 y ← p[x]
4 while y ≠ NIL and x = right[y]
5
do x ← y
6
y ← p[y]
7 return y
• if the right subtree of node x is empty and x has
a successor y, then y is the lowest ancestor of x
whose left child is also an ancestor of x.
Insertion
TREE-INSERT(T, z)
1 y ← NIL
2 x ← root[T]
3 while x ≠ NIL
4
do y ← x
5
if key[z] < key[x]
6
then x ← left[x]
7
else x ← right[x]
8 p[z] ← y
9 if y = NIL
10
then root[T] ← z
11 else if key[z] < key[y]
12
then left[y] ← z
13 else right[y] ← z
// Tree T was empty
Deletion
Deletion
Which node is actually
removed depends on how
many children z has.
(a) If z has no children, we
just remove it.
(b) If z has only one child,
we splice out z.
(c) If z has two children, we
splice out its successor
y, which has at most one
child, and then replace
z's key and satellite data
with y's key and satellite
data.
Sorting and BST
• if there is an array A, can we sort this
array using binary search tree operations
as a black box?
– Build the binary search tree, and then
traverse it in order.
Sorting and BST
• Example: A= [3 1 8 2 6 7 5], what we will
get?
• What's the running time of the algorithm?
Cost of build a BST of n nodes
• Can anybody guess an answer?
– Θ(nlgn)
• in most case
– Θ(n2)
• in the worst case
• Does this looks familiar and remind you of
any algorithm we've seen before?
– Quicksort
• Process of Quicksort
BST sort and Quicksort
• BST sort performs the same comparisons
as Quicksort, but in a different order!
• So, the expected time to build the tree is
asymptotically the same as the running
time of Quicksort.
Informal Proof
• The depth of a node equals to the number of
comparisons made during TREE-INSERT.
• Assuming all input permutations are equally
likely, we have
Average node depth()
=
(comparison times to insert node i)
= Θ(nlgn)
= Θ(lgn)
• The expected running time of quicksort on
n elements is?
Expected tree height
• Average node depth of a randomly built
BST = Θ(lgn).
• Does this necessarily means that its
expected height is also Θ(lgn)?
Outline of Formal Proof
• Prove Jensen’s inequality, which says that
f(E[X]) ≤ E[f(X)] for any convex function f
and random variable X.
• Analyze the exponential height of a
randomly built BST on n nodes, which is
the random variable Yn= 2Xn, where Xn is
the random variable denoting the height of
the BST.
• Prove that 2E[Xn] ≤ E[2Xn] = E[Yn] = O(n3),
and hence that E[Xn] = O(lgn).
Convex functions
• A function f is convex if for all α, β≥0 such
that α+β=1, we have
f(αx+ βy) ≤αf(x)+βf(y)
for all x,y∈R.
Convexity lemma
Lemma.
• Let f be a convex function, and let α1, α2 ,
…, αn be nonnegative real numbers such
that
. Then, for any real numbers
x1, x2, …, xn, we have
• How to prove that?
Proof of convexity lemma
Proof.
• By induction on n. For n=1, we have α1=1
and hence f(α1x1) ≤α1f(x1) trivially.
• Inductive step:
why we introduce 1-αn?
By convexity
By induction
Convexity lemma: infinite case
Lemma.
• Let f be a convex function, and let α1, α2 ,
… be nonnegative real numbers such that
. Then, for any real numbers x1,
x2, … , we have
assuming that these summations exist.
• We will not prove that, but intuitively its
true.
Jensen’s inequality
• Jensen’s inequality, which says that
f(E[X]) ≤ E[f(X)] for any convex function f
and random variable X.
Proof.
By definition of expectation
By Convexity lemma (infinite case)
Analysis of BST height
• Let Xn be the random variable denoting
the height of a randomly built binary
search tree on n nodes, and let Yn= 2Xn be
its exponential height.
• If the root of the tree has rank k, then
Xn= 1 + max{Xk–1,Xn–k}
since each of the left and right subtrees
are randomly built. Hence, we have
Yn= 2·max{Yk–1,Yn–k}.
Analysis (continued)
• Define the indicator random variable Znk as
• Thus, Pr{Znk= 1} = E[Znk] = 1/n, and
Take expectation of both sides.
Analysis (continued)
Linearity of expectation.
Independence of the rank of the root from the ranks of
subtree roots.
The max of two nonnegative numbers is at most their
Each term appears
sum, and E[Znk] = 1/n.
twice, and reindex.
Analysis (continued)
• How to solve this?
– Substitution Method
• Guess E[Yn] ≤ cn3 for some positive
constant c.
• Initial conditions: for n=1, E[Y1]=2X1=2≤c,
we can pick c sufficiently large to handle.
Analysis (continued)
• Substitution:
• How to compute the expression on right?
– Integral method.
The grand finale
• By Jensen’s inequality, since f(x) = 2x is
convex, we have
What we just showed
• Taking lg of both sides yields
Post mortem
• Why bother with analyzing exponential
height?
• Why not just develop the recurrence on
Xn= 1 + max{Xk–1,Xn–k}
directly?
Answer
• The inequality max{a,b} ≤ a+b provides a
poor upper bound, since the RHS
approaches the LHS slowly as |a–b|
increases.
• The bound
allows the RHS to approach the LHS far
more quickly as |a–b| increases. By using
the convexity of f(x) = 2x via Jensen’s
inequality, we can manipulate the sum of
exponentials, resulting in a tight analysis.