A-level Computing
COMP3 Tree Traversals
COMP3: Section 3.3.2 Programming Concepts: Standard Algorithms
Graphs and trees
In this topic you will cover:
• rooted trees
• multiway trees
• binary trees
• in-order, pre-order and post-order tree traversals of a binary tree.
Rooted trees
A rooted tree is an abstract data type consisting of nodes arranged in hierarchical fashion, except
when the tree is empty, linked by branches with a single node called the root at the top of the tree
and every other node a descendent of the root. All nodes below the root have one and only one
parent node as shown in Figure 1.
A node that has a parent and itself is a parent of one or more nodes is known as an internal node.
A node that has a parent node but is not a parent of other nodes is known as a leaf or terminal
node.
Figure 1
An example of a multiway tree
The root node
An internal node
A leaf node
A node

A branch
See also Section 2.5 of the Nelson Thornes book, AQA Computing A2, by Kevin Bond & Sylvia Langfield for more
information.
Choose option
A-level Computing – COMP3 – Tree traversals
Binary Trees
A binary tree is a tree in which each node is the parent of at most two other nodes is known as a
binary tree as shown in Figure 2.
In a binary tree, there are never more than two branches descending from a node.
Figure 2
An example of a binary tree
The root node
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A-level Computing – COMP3 – Tree traversals
An alternative definition of a binary tree is as follows.
A binary tree is an abstract data type defining a finite set of elements. The set is either empty
or is partitioned into three subsets. The first subset contains a single element called the
root of the tree. The other two subsets are themselves binary trees, called the left and right
sub-trees of the original tree.
This alternative view of a tree is shown in Figure 3.
Figure 3
Root, left sub-tree and right sub-tree view of binary tree
The root node
Left sub-tree
Right sub-tree
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A-level Computing – COMP3 – Tree traversals
This view of the original binary tree can be applied to each sub-tree. For example the left sub-tree
consists of a root and two sub-trees. In turn each of these new sub-trees can be defined in terms of
their root, left and right sub-trees, et cetera. Eventually, the sub-trees so defined satisfy the empty
case. Figure 4 illustrates this in a simpler example.
Figure 4
Division of a binary tree into
Root of
left subtree
a root, left sub-tree and right sub-tree
Root
Left
sub-tree
Left
sub-sub-tree
which happens
to be empty
Right
sub-tree
Right
sub-sub-tree which
happens to be empty
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A-level Computing – COMP3 – Tree traversals
Why binary trees are useful
Binary trees are useful because they can represent the order in which mathematical expressions
are evaluated. For example, the mathematical expression
(z + 6) * (y –3)
is represented in a binary tree as follows:
Figure 5
An evaluation tree
*

+
z
6
y
3
The expression in the tree is evaluated from the bottom up and from left to right.
Table 1 shows the order of evaluation.
Table 1
z+6
y+3
intermediate result 1 * intermediate result 2
Evaluation order
intermediate result 1
intermediate result 2
final result
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A-level Computing – COMP3 – Tree traversals
Binary Tree as a Data Structure
Processing a binary tree on paper is straightforward. We have just done so with the evaluation tree
in Figure 5.
Processing a binary tree by computer is not so straightforward when the programming language
that is used does not have a binary tree data type. Compare this situation with processing integer
values in Pascal. Pascal has a native data type called Integer and in-built operators that act on
integer values and variables of the integer data type. On the other hand, Pascal does not have a
native data type to represent a binary tree nor does it have any in-built operators for manipulating
binary trees.
Using Pascal for binary tree processing thus requires the programmer to build binary trees from
what is available in the language. The building blocks that are used are called the primitives of
the language.
We shall use the array primitive and the record primitive to construct a data structure in Pascal that
represents a binary tree. The binary tree in Figure 5 now becomes a data structure in which the
nodes are records with the field structure:
NodeType = Record
LeftPointer : Integer;
Item : String;
RightPointer : Integer;
End;
A pointer is a kind of address.
Figure 6 illustrates this. Each record is numbered (given an address). The record numbers range
from 1 to 7. Zero is used to represent the null pointer, the pointer that points nowhere.
Figure 6
Binary tree represented by a collection of records
Item
1
LeftPointer
2
*
RightPointer
3
2
4
+
3
5
4
0
z
6
5
0
0
6
-
7
6
1
0
y
7
0
0
3
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A-level Computing – COMP3 – Tree traversals
The records need to be stored in a one-dimensional array with the structure
ArrayOfNodesType = Array[1..7] Of NodeType;
A variable called Root node pointer stores a pointer to the root node row of the array.
Figure 7
Root node
pointer
1
Binary tree stored in an array of records
1
2
*
3
2
4
+
5
3
6
-
7
4
0
z
0
5
0
6
0
6
0
y
0
7
0
3
0
Array subscript
Ordered Binary Tree
It is possible using a binary tree to sort an unordered list of items into a particular order. For
example, consider the following unordered list of integers:
30 20 41 25 37 17 56
An ordered binary tree is created as follows. Starting from the leftmost integer in the list, the root
node is created first.
30
The next integer is taken and placed in the left sub-tree if it is numerically less than the root node
integer (or equal) or in the right sub-tree if it is numerically greater.
30
20
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A-level Computing – COMP3 – Tree traversals
The next integer is numerically greater than the root node integer and so goes into the right subtree.
30
20
41
The next integer is numerically less than the root node integer but greater than the integer at the
root of the left sub-tree so it is placed to the right of this node.
30
20
41
25
The remainder of the list is processed to produce the finished ordered binary tree shown in Figure
8.
Figure 8
An ordered binary tree
30
20
17
41
25
37
56
By visiting the nodes of this tree in left to right, bottom to top order an ordered equivalent of the
original list is produced. This ordered list has placed the integers in ascending numerical order as
follows:
17 20 25 30 37 41 56
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A-level Computing – COMP3 – Tree traversals
Traversal of a binary tree
Traversal of a binary tree means to visit each node of the tree in some order.
Usually, the visit means do something with the node, e.g. print its data value.
There are three principle ways of traversing a binary tree:
1. Pre-order traversal
2. In-order traversal
3. Post-order traversal
These traversals can be specified recursively. The trick is to realise that a binary tree can be
defined recursively. The recursive definition states:
a tree is either empty or consists of
1. A root
2. A left sub-tree
3. A right- sub-tree.
Algorithms
Pre-order Traversal
a. If tree empty do nothing
Otherwise
b. Visit the root (e.g. print the value of the data item at the root)
c. Traverse the left sub-tree in pre-order
d. Traverse the right sub-tree in pre-order
In-order Traversal
1. If tree empty do nothing
Otherwise
2. Traverse the left sub-tree in in-order
3. Visit the root
4. Traverse the right sub-tree in in-order
Post-order Traversal
1. If tree empty do nothing
Otherwise
2. Traverse the left sub-tree in post-order
3. Traverse the right sub-tree in post-order
4. Visit the root
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A-level Computing – COMP3 – Tree traversals
Examples
D
a. pre-order DAE
b. in-order ADE
c. post-order AED
A
E
D
B
A
F
C
E
G
a. pre-order DBACFEG
b. in-order ABCDEFG
c. post-order ACBEGFD
+
*
A
/
C
E
G
a. pre-order +*AC/EG
b. in-order A*C+E/G (order of evaluation is (A*C) + (E/G))
c. post-order AC*EG/+ (Reverse Polish form)
Tracing a particular traversal of a binary tree
There are two ways to help you trace a particular traversal of a binary tree.
First method
The first places labels on nodes to aid remembering how far the trace has reached. Figure 9 shows
a partial trace for a pre-order traversal through the given tree. All three stages of the algorithm
have been applied to the left sub-tree and the root. The right sub-tree is being traced currently.
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A-level Computing – COMP3 – Tree traversals
Figure 9
Partial tracing of binary tree
If tree empty do nothing
Otherwise
a. Visit the root (e.g. print the value of the data item
at the root)
b. Traverse the left sub-tree in pre-order
c. Traverse the right sub-tree in pre-order
ab
D
abc
a
A
E
Second method
This involves drawing an outline around the tree for each traversal as follows:
Pre-order traversal
Starting to the left of the root, draw an outline around the tree as shown in Figure 10. As you pass
to the left of a node (where the red dot is marked) output the data in that node, i.e. DBACFEG
Figure 10
Pre-order traversal of a binary tree using an outline around the tree as a guide
D
B
A
F
C
E
G
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A-level Computing – COMP3 – Tree traversals
In-order traversal
Starting to the left of the root, draw an outline around the tree as shown in Figure 11. As you pass
underneath a node (where the blue dot is marked) output the data in that node, i.e. ABCDEFG
Figure 11
In-order traversal of a binary tree using an outline around the tree as a guide
D
B
A
F
C
E
G
Post-order traversal
Starting to the left of the root, draw an outline around the tree as shown in Figure 12. As you pass
to the right of a node (where the green dot is marked) output the data in that node, i.e. ACBEGFD
Figure 12
Post-order traversal of a binary tree using an outline around the tree as a
guide
D
B
A
F
C
E
G
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A-level Computing – COMP3 – Tree traversals
End of sub-topic questions
1
What output is produced by each of the following tree traversals applied to the binary tree in
Figure 13?
(a)
(b)
(c)
Pre-order
In-order
Post-order
Figure 13
Binary tree
+
*
/
-
A
E
+
B
C
G
D
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A-level Computing – COMP3 – Tree traversals
2 (a)
An algebraic expression is represented in a binary tree as follows.
*
+
A
(i)

B
C
D
Using a copy of this tree, label its root, a branch and a leaf node.
(3 marks)
(ii) On the same copy, mark and label the left sub-tree and the right sub-tree
of this
tree.
(2 marks)
(b) A recursively-defined procedure T, which takes a tree structure, tree(x, y, z) as
its
single parameter, where x is the root, y is the left sub-tree and z is the right
sub-tree, is
defined below. (<> means not equal to)
Procedure T ( tree(x, y, z) )
If y <> empty
Then
PRINT ‘(‘
T(y)
EndIf
PRINT x
If z <> empty
Then
T( z )
PRINT ‘)’
EndIf
EndProc
What is meant by recursively-defined?
(1 mark)
(c)
Explain why a stack is necessary in order to execute procedure T recursively.
(3 marks)
(d)
Dry run the following procedure call
T(
tree( ‘*’, tree ( ‘+’, tree ( ‘A’, empty, empty ), tree ( ‘B’, empty, empty ) ),
tree ( ‘-’, tree ( ‘C’, empty, empty ), tree ( ‘D’, empty, empty ) )
)
)
showing clearly the PRINTed output and the values of the parameter omitted
table (rows 4, 5, 6, 7) for the seven calls of T.
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A-level Computing – COMP3 – Tree traversals
Call
Number
Parameter
1
tree(‘*’, tree(‘+’, tree(‘A’,empty,empty), tree(‘B’,empty,empty) ),
tree(‘-’, tree(‘C’,empty,empty), tree(‘D’,empty,empty) )
)
2
tree(‘+’, tree(‘A’,empty,empty), tree(‘B’,empty,empty))
3
tree(‘A’,empty,empty)
4
5
6
7
(10 marks)
(e)
What tree traversal algorithm does procedure T describe?
3
The algebraic expression
(1 mark)
A*B+C*D
as
is stored as a binary tree in the three arrays term, leftpointer and rightpointer,
shown in the table.
Term
LeftPointer
RightPointer
(a)
1
+
2
3
2
*
4
5
Subscript
3
4
5
*
A B
6
0
0
7
0
0
6
C
0
0
7
D
0
0
What is a binary tree?
Draw a diagram of the binary tree represented by the three arrays term,
rightpointer.
leftpointer,
(6 marks)
(b) A recursively-defined procedure P, which takes an integer as its single parameter, is defined
below.
Procedure P (Subscript)
If LeftPointer[Subscript] > 0
Then P (LeftPointer[Subscript])
EndIf
If RightPointer[Subscript] > 0
Then P (RightPointer[Subscript])
EndIf
Print Term[Subscript]
End Of P
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A-level Computing – COMP3 – Tree traversals
What is meant by recursively-defined?
(c)
(1 mark)
Explain why a stack is necessary in order to execute procedure P recursively.
(2 marks)
(d)
Using as an aid a copy of the partially completed table shown below, and given
that
the three arrays term, leftpointer and rightpointer are global, dry run the
procedure call P (1)
showing clearly the PRINTed output and the values of the
parameter omitted from the table
for the seven calls of P.
------------------------|
Call
| Parameter |
|
Number |
|
-------------------------|
1
|
1
|
|
2
|
2
|
|
3
|
4
|
|
4
|
5
|
|
5
|
|
|
6
|
|
|
7
|
|
------------------------(10 marks)
(e)
What tree traversal algorithm does procedure P describe?
(1 mark)
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A-level Computing – COMP3 – Tree traversals
Solutions
1
(a) pre-order: +*-AB+CD/EG
(b) in-order: (A-B)*(C+D)+(E/G)
(c) post-order: AB-CD+*EG/+
(3 marks)
2 (a)(i)
Root
*
-
+
A
B
Branch
Labelling must
clearly indicate term
D
C
Leaf
Node
(3 marks)
(a)(ii)
Left
Sub Tree
Right
Sub Tree
*
+
A
Must clearly indicate
subsets

B
C
D
(2 marks)
(b) A procedure which is defined in terms of/calls itself/re-entrant
(1 mark)
(c) State of machine/return address/parameter
needs to be stored/held
to enable a previous execution of T to be resumed
Or
so that each call to T
can pass
a new value of the parameter
(3 marks)
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A-level Computing – COMP3 – Tree traversals
(d)
Call
Number
Parameter
1
tree(‘*’, tree(‘+’, tree(‘A’,empty,empty), tree(‘B’,empty,empty) ),
tree(‘-’, tree(‘C’,empty,empty), tree(‘D’,empty,empty) )
)
2
tree(‘+’, tree(‘A’,empty,empty), tree(‘B’,empty,empty))
3
tree(‘A’,empty,empty)
4
tree(‘B’,empty,empty)
1
5
tree(‘-’,tree(‘C’,empty,empty),tree(‘D’,empty,empty) )
1
6
tree(‘C’,empty,empty)
1
7
tree(‘D’,empty,empty)
1
1
1
1
( ( A +B ) * ( C – D ) )
1
1
Must be in
middle for mark
1
(max 10 marks)
(e) In-order traversal
(1 mark)
(Total: 20 marks)
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A-level Computing – COMP3 – Tree traversals
3 (a) 1 mark for stating what constitutes a binary tree:
A tree in which each node is the parent of at most two other nodes is known as a
binary tree.
Or
A finite set of elements which is either empty or is partitioned into three disjoint
subsets. The first subset contains a single element called the root of the tree. The
other two subsets are themselves binary trees, called the left and right subtrees of
the original tree.
Or
A diagram showing more than two levels
Or
A labelled diagram showing just two levels and using terms root|root node,
node|leaf|leaf node.
(1 mark)
-------N.B. accept nodes without operators but with subscripts which correctly reference these operators
+
*
A
(1 mark)
(1 mark)
(1 mark)
B
*
(1 mark)
C
(1 mark)
D
(max 6 marks)
(b) Procedure which is defined in terms of itself, i.e. contains within its body a
reference to itself.
Or
A procedure which calls itself.
(1 mark)
(c)The current state of the machine/one of: return address, procedure parameter, status
register, other register values, local variables - must be saved/preserved
so can return correctly to previous invocation of P
(1 mark)
Or
because P must be re-entrant
(2 marks)
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A-level Computing – COMP3 – Tree traversals
(d) N.B. Accept printed output in form AB*CD*+ Or
A
B
*
C
D
*
+
...1
...1
...1
...1
...1
...1
...1
For those candidates who make a mistake the following guide should be followed: marks
are for
A first(1),
if A not first AB (1)|AB*(2)
B in wrong place B*(1)
in any position CD(2)|CD*(3)
in wrong position D*(1)|D*+(2)
e.g. CD*AB*+ gets a mark of 6
(max 7 marks)
Parameter values @ 1 mark each up to max 3 marks
Call No
1
2
3
4
5
6
7
Parameter Printed Output
1
+
2
*
4
A
5
B
3
..1 mark
*
6
..1 mark
C
7
..1 mark
D
(e) Post-order/converts infix to postfix
(1 mark)
(Total 17 marks)
In this topic you have covered:
• a rooted multiway tree
• a binary tree
• why binary trees are useful
• binary tree as a data structure
• constructing an ordered binary tree
• tree traversals - pre-order, in-order, post-order
• tracing tree traversals.
Reproduced by permission of Dr K Bond. All rights reserved.
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