More on Trees
University
Fac. of
Sci. & Eng.
CS Dept.
EE Dept.
Bus. School
Law School
Math.
Dept.
Trees
1
Linked Structure for Trees
A node is represented by
an object storing



Element
Parent node
Sequence of children
nodes

B

Node objects implement
the Position ADT
A
B
D
A
C

D
F
F

E
C
Trees

E
2
Linked Structure for Binary Trees
A node is represented by
an object storing





Element
Parent node
Left child node
Right child node
B
Node objects implement
the Position ADT

B
A

A
D
C
D

E

C
Trees


E
3
Linked Structure for Binary Trees
A node is represented by
an object storing





Element
Parent node
Left child node
Right child node
B
Node objects implement
the Position ADT

B
A

A
D

D
C
F
D


E
C
G
Trees




4
G

Preorder Traversal
A traversal visits the nodes of a
tree in a systematic manner
In a preorder traversal, a node is
visited before its descendants
Application: print a structured
document
1
Algorithm preOrder(v)
visit(v)
for each child w of v
preorder (w)
Make Money Fast!
2
5
1. Motivations
9
2. Methods
3
4
1.1 Greed
1.2 Avidity
6
7
2.1 Stock
Fraud
Trees
2.2 Ponzi
Scheme
References
8
2.3 Bank
Robbery
5
Preorder Traversal (Another
example)
My Explanation:
When a node is reached,
visit it.
Visit the sub-trees rooted
by its children one by one.
Algorithm preOrder(v)
visit(v)
for each child w of v
preorder (w)
1
2
3
4
9
6
5 7
17
14
10
8 11
13
Trees
12
15
166
Postorder Traversal
In a postorder traversal, a
node is visited after its
descendants
Application: compute space
used by files in a directory and
its subdirectories
9
Algorithm postOrder(v)
for each child w of v
postOrder (w)
visit(v)
cs16/
3
8
7
homeworks/
todo.txt
1K
programs/
1
2
h1c.doc
3K
h1nc.doc
2K
4
5
DDR.java
10K
Trees
Stocks.java
25K
6
Robot.java
20K
7
Postorder Traversal (Another
example)
My explanation:
Algorithm postOrder(v)
for each child w of v
postOrder (w)
visit(v)
. If the reached node is a leaf,
then visit it.
When a node is visited, visit the
sub-tree rooted by its sibling on
the right.
When the leftmost child is visited,
visit its parent.
17
7
3
1
15
14
6
2 4
16
10
5 8
11
Trees
9
12
138
Evaluate Arithmetic Expressions
Specialization of a postorder
traversal


recursive method returning
the value of a subtree
when visiting an internal
node, combine the values
of the subtrees
+

Algorithm evalExpr(v)
if isExternal (v)
return v.element ()
else
x  evalExpr(leftChild (v))
y  evalExpr(rightChild (v))
  operator stored at v
return x  y

-
2
5
3
2
1
Trees
9
Array-Based Representation of
Binary Trees
nodes are stored in an array
1
A
…
2
3
B

D
let rank(node) be defined as follows:



4
rank(root) = 1
if node is the left child of parent(node),
rank(node) = 2*rank(parent(node))
if node is the right child of parent(node),
rank(node) = 2*rank(parent(node))+1
Trees
5
E
6
7
C
F
10
J
11
G
H
10
Full Binary Tree
A full binary tree:
 All the leaves are at the bottom level
 All nodes which are not at the bottom level have two
children.
 A full binary of height h has 2h leaves and 2h-1 internal
nodes.
1
3
2
4
5
6
7
A full binary tree of height 2
This is not a full binary tree.
Trees
11
Properties of Proper Binary Trees
Notation
n number of nodes
e number of
external nodes
i number of internal
nodes
h height
1
2
Properties for proper
binary tree:
 e = i + 1
 n = 2e - 1
 h  i
 e  2h
3  h  log2 e
7
6
14
Trees
15
No need to
remember.
12
Depth(v): no. of ancestors of v
Algorithm depth(T,v)
If T.isRoot(v) then return 0;
else
return 1+depth(T, T.parent(v))
0
Make Money Fast!
1
1
1. Motivations
1
2. Methods
2
2
1.1 Greed
1.2 Avidity
2
2
2.1 Stock
Fraud
Trees
2.2 Ponzi
Scheme
References
2
2.3 Bank
Robbery
13
Height(T): the height of T is max
depth of an external node
Algorithm height1(T)
h=0;
for each vT.positions() do
if T.isExternal(v) then h=max(h, depth(T, v))
return h
T.positions() holds all nodes in T. See the method
in LinkedBinaryTree.java. The max is 2.
0
1
2
Make Money Fast!
2. Methods
1. Motivations
1.1 Greed
1.2 Avidity
2
2.1 Stock
Fraud
Trees
2
2.2 Ponzi
Scheme
2
1
2.3 Bank
Robbery
2
14
Height(T,v):
1. If v is an external node, then height of v is 0.
2. Otherwise, the height of v is one +max height of
a child of v.
Algorithm height2(T,v)
if T.isExternal(v) then return 0
else
h=0
for each wT.children(v) do
h=max(h, height2(T, w))
return 1+h
Trees
15
Height(T,v):
Algorithm height2(T,v)
if T.isExternal(v) then return 0
else
h=0
for each wT.children(v) do
h=max(h, height2(T, w))
return 1+h
2
Make Money Fast!
1
1
1. Motivations
1
2. Methods
0
0
1.1 Greed
1.2 Avidity
0
0
2.1 Stock
Fraud
Trees
2.2 Ponzi
Scheme
References
0
2.3 Bank
Robbery
16
Next time: Chapter 7. Heap
Exercise1: Given a binary tree T, and two nodes u and v
in T, test if u is an ancestor of v.
Boolean isAncestor(T, u, v)
Exercise2: Suppose that we use array-based
representation for a binary tree. Give the positions of
the nodes in the following tree
+


-
2
5
3
2
This time is slightly worse than last time. I
should prepare the speech carefully. 5.33?
1
Trees
17