COP 3540 Data Structures with OOP
Chapter 8 - Part 1
Binary Trees
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Why Trees?
 Trees are one of the fundamental data structures.
 Many real-world phenomena cannot be
represented w/data structures we’ve had so far.
 Think of arrays:
 Easy to search, especially if ordered.
• O(log2n) performance! Great. (binary search)
 Inserting; Deleting? Horrible if ordered. Must
find item or place before actions.
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Why Trees?
 How about Linked Lists?
 Inserts and deletes? Great. Take O(1) time –
the best you can get! (if inserting / deleting
from one end)
 Searching? Search to insert / delete/ change?
Not nearly as good as O(1) or even O(log2n)!
• On average, must search n/2 items!
• Process requires O(n) time.
• Ordering the linked list may help, as we
must still search to find.
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Trees
 First, trees in general.
 Consists of nodes connected by edges.
 Trees have indegree <=1, no cycles;
• Implies one ‘path’ to a node.
 Trees with indegree > 1 and cycles = graphs.
• More later on graphs.
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Trees
 General Tree
A
root
node
B
D
C
edge
F
E
G
H
Edges connect nodes.
Only way to get to one node is along an indicated path (edges) and these are downward.
Edges are represented in a program by ‘references;’ nodes as likely primitives or objects.
One node at top of tree called the root. Can only have one root.
Binary Trees have ‘outdegree’ <= 2 for ALL nodes.
Multi-way tree can have outdegree >= 2 for at5 least one node (see above).
Trees
 General Tree
A
root
node
B
D
C
edge
F
E
G
H
Parent: All nodes have exactly one parent.
Child: Any node may have one or more lines coming from it: children
Binary tree has at most two children emanating from a node.
Leaf: node with no children
Subtree: any node may be considered to be a root of a subtree, even leaves.
Visiting: term used to indicate that a node is visited under program control – usually to
process the data at that node. Merely passing over a node does not constitute a visit.
Traversing: refers to visiting all nodes in a tree in a prescribed manner.
Levels: start with level 0.
Keys: normally what is displayed on the tree,
like A, B, C, … above.
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Trees
 Binary Tree
A
root
node
B
D
edge
F
E
G
H
Binary Tree: every node has no more than two children.
A child node is called a left child or right child, but may, in turn, be the root of a subtree!
A node in a binary tree may have no children.
We normally talk in terms of binary search trees.
In theory (logical; abstraction), can exist to any number of levels;
in practice (i.e. implementation), can run out of memory space.
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How Do Binary Search Trees Work?
 Need to carry out basic tree operations
such as




finding a node,
traversing a tree (get around in the tree),
adding a node,
deleting a node, etc.
 This is what this chapter is all about.
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Unbalanced Trees
Note: tree is ‘unbalanced.’
root
This means nodes are mostly on
one side or the other.
Tree may be nearly balanced until certain levels
Then, it may become quite unbalanced. Skewed.
A
D
Trees can become unbalanced due to the way
they were created.
Generally, they are more balanced, if randomly developed.
F
H
Greatly prefer balanced trees.
Have very welcome properties
Unbalanced trees present problems in efficiently processing them.
Red-Black trees address unbalanced trees (further ahead).
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G
Trees in Java Code
 So, how do we implement binary trees in Java?
 Normally, we will store the nodes at unrelated places
in memory with references to children as we are
accustomed to do.
 Can also represent a tree in memory as an array, with
children located in specific positions within this array.
 Will look later at this.
 Let’s look at some code segments now.
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The Node Class
We need a class of node objects.
These will contain appropriate data and up to two references
to children (can have MORE, as we shall see later…)
class Node
{
int iData;
// data used as key value
double fData
// other data
Node leftChild;
// this node’s left child
Node rightChild;
// this node’s right child.
public void displayNode()
{
// whatever…
}// end display()
} // end class Node.
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Sometimes the data might be objects rather than primitives,
and better simply referenced in the Node itself.
class Node
{
Person p;
Node leftChild;
Node rightChild;
// reference to a person object
// this node’s left child
// this node’s right child.
public void displayNode()
{
// whatever…
}// end display()
} // end class Node.
class Person
{
int iData;
double fData; …
} // end class person
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The Tree Class
 Need a tree class from which a tree object can be
instantiated.
 Will call class Tree with one field: a Node variable
that references the root.
 Identical to ‘first’ and ‘last’ for linked lists…(get started)
 Also, since we do not allocate the entire structure at
one time (like an array), we can ONLY have a pointer
to the first element or root.
 Consider the basic format of a Tree class:
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Tree Class
class Tree
{
private Node root;
// only data field in Tree; but key!
public void find (int key)
{
// stub: not showing details of this method here
}// end find()
public void insert (int id, double dd)
{
// stub; placeholder
}// end insert()
pubic void delete (int id)
{
// stub
}// end delete()
} // end class Tree.
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The TreeApp Class
class TreeApp
{
public static void main (String [ ] args)
{
Tree theTree = new Tree; // make a tree.
// creates an object of type Tree. (previous slide)
theTree.insert (50, 1.5);
theTree.insert(25, 1.7);
theTree.insert(75, 1.9);
//insert three nodes
// invoking tree methods…
// no implementations shown yet.
Node found = theTree.find(25); // find node with key 25
if (found != null)
// So what does this do???
System.out.println (“Found the node with key 25”);
else
System.out.pirntln(“ Could not find node with key 25”);
// end main()
} // end class TreeApp.
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Java Code for Finding a Node
 Insert and delete a bit later.
 Start with finding a node
 Remember, nodes have values.
 In building a binary search tree, we ‘assume’
nodes are built in an order; that is the values to
the left of the root are smaller than the parent or
root, while values to the right are larger.
 Assume the binary search tree is already built:
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Sample Binary Tree
 Note the left and right, less than, greater
than relationships.
50
30
20
60
40
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Discuss the code:
public Node find (int key)
{
Node current = root;
// assumes non-empty tree
// start at root
while (current.iData != key)
// if no match
{
if (key < current.iData)
current = current.leftChild;
// recall: current = current.next??
else
current = current.rightChild;
If (current == null)
return null;
// not found; boundary condition
} // end while
return current;
// returns reference to node
} // end find()
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Java Code for Inserting a Node
 Must find the place where to insert new node.
 Follow a path to parent where we can insert child
 Connect new Node to parent as left or right child
 This depends on whether the new node is greater
than or less than the value of the parent.
 First: create a new node.
 Then, use similar to ‘find’ to locate new node.
 Ignore dupes at this time
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public void insert(int id, double dd) // we are within TreeApp.java file…
{
Node newNode = new Node(); // make new node
newNode.iData = id;
// insert data
// create new node; move in its data
newNode.dData = dd;
if(root==null)
// no node in root.
root = newNode;
// if true, we are done.
else
// root occupied
{
// else not root
Node current = root;
// start at root Current is a pointer to a node and it points to root.
Node parent;
// creating a reference to a parent (think singly-linked list!!! Needed ‘previous’)
while(true)
// (exits internally)
// here is our search to find the right spot.
{
parent = current;
if (id < current.iData) // go left?
{
current = current.leftChild; // But maybe there IS no left child. So:
if(current == null)
// means there is no left child of current
{
parent.leftChild = newNode;
// insert on left . Link this new node in. We are done.
return;
// we are done if we get here.
}// end if
else
// go right
// current > current.iData? If so, go right?
{
current = current.rightChild;
if(current == null)
{
// if end of the line insert on right
parent.rightChild = newNode;
// link in and we are done again.
return;
// we are done if we get here.
}// end if
} // end else
} // end while
}// end else not root
Note: we use ‘parent’ to keep track of where we are…
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} // end insert()
Parent is used to keep track of the last non-null node.
Traversing the Tree
 Traversing the tree means ‘visiting’ all the
nodes is some kind of specified order.
 Not particularly fast (unless you use recursion).
 Three basic ways (there are several others…)
 Preorder traversal (NLR scan or traversal)
 Inorder traversal (LNR traversal)
 Postorder traversal (LRN traversal)
 Most common is inorder.
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Inorder Traversal – Binary Tree (LNR Traversal)
 Results in ascending scan based on key values.
 Simplest way to traverse a tree is via recursion
 Start with ‘a’ node as an argument (start with root)
 Recursive routine will
 Call itself to traverse the node’s left subtree
 Visit the node (implies do something with it!)
 Call itself to traverse the node’s right subtree.
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Java Code for Traversing – Inorder Recursive…
 Traversals typically take a total of three statements,
if executed recursively. Here’s the LNR traversal:
private void inOrder (Node localRoot)  initially called with root,
{
// as in inOrder(root);
if (localRoot != null)
{
inOrder (localRoot.leftChild)
System.out.print(“localRoot.iData + “ “);
inOrder (localRoot.rightChild);
}// end if
} // end inOrder()
It continues until there are no more nodes to visit. All nodes visited!
Displays value of the data via System.out.print statement.
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Execute the algorithm: LNR Traversal (inOrder)
private void inOrder (node localRoot)
{
(initially called with root as in inOrder(root);)
if (localRoot != null)
{
inOrder (localRoot.leftChild)
System.out.print(“localRoot.iData + “ “);
inOrder (localRoot.rightChild);
}// end if
} // end inOrder() // inOrder means priority Left!
Start with 50 (root) - Check left subchild L; not null
Recursive call with 30 as root; check with left subchild L; not null; 20
Recursive call with 20 as root; check with left subchild; L null.
Visit 20 (print) N (System.out.print above…)
50
30
60
40
Recursive call with right subchild; R null – completed most recent call to the left
Visit 30 (print) N (System.out.print above…)
Recursive call with right subchild (40)
Recursive call with left subchild of 40; It is null L
Visit 40 (print) N
Recursive call with right subchild of 40. It is null – completed another call on right
Visit 50 N (System.out.print above)
Recursive call with right subtree (60) (not null)
Recursive call with left subtree of N (60) It is null L
Visit 60.
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Recursive call with right subtree of N (60) It is null Done!
Priority: L-N-R
 Inorder traversal (LNR) means
Left. Go to the left if at all possible; Continue going to the
left as much as possible. THEN:
process that node
 Node Next, the root (node) N; process it
 Right. Lastly, go to the right, R. But then try to go left if at
all possible, then N, and process it, then Right… process
the Right if there is nothing to t left. Note: the Right then
becomes N.
 But even when you go to the right, before you process it,
you must determine if you can go to the left again, as
much as possible…recursively…
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 Inorder, preorder, postorder get their names
from position of N in LNR, NLR, and LRN.
 In traversing the tree, the first letter in the
scan is the priority, second letter is second
priority, and third letter is last priority.
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Pre-Order (NLR) and Post-Order (LRN) Traversals
 NLR and LRN traversals. Very important.
 Same statements in algorithm: Change order!!
 NLR, visit first and then go left, which would be
considered the node, N, of the left subtree (if present
and not null).
 Process this node (visit the node).
 Go left. If not null, this node is now the new root, N,
of another subtree. We process that…etc.
 Simple priority: N – L – R and
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Can see only the order of recursive calls is changed.
 inOrder (localRoot.leftChild) // LNR recursive scan
inOrder (localRoot.leftChild)
System.out.print(“localRoot.iData + “ “);

inOrder (localRoot.rightChild);
 preOrder (localRoot.leftChild) // NLR recursive scan
System.out.print(“localRoot.iData + “ “);
inOrder (localRoot.leftChild)
inOrder (localRoot.rightChild);
 postOrder (localRoot.leftChild) // LRN recursive scan
inOrder (localRoot.leftChild)
inOrder (localRoot.rightChild);
System.out.print(“localRoot.iData + “ “);

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Pre-Order and Post-order Traversals
 Are specific applications for these traversals.
 A binary tree (not a binary search tree) can be
used to represent algebraic expressions that
involve binary operators.
 The root holds the operator and the other nodes
hold a variable (operands) or another operator.
 Each subtree is a valid algebraic expression.
 Consider the following:
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Pre-Order Traversal:
(NLR)
This represents the expression (A+B)*C
This is called ‘infix’ notation – which we are used to.
*
+
C
For preorder traversals, NLR, we have the algorithm:
visit the node
A
B
call itself to traverse the node’s left subtree
call itself to traverse the nodes’s right subtree See the priority???
Above, preorder would be: *+ABC
This is also called ‘prefix’ notation. (sometimes called Polish prefix)
Advantage: parentheses are never required.
Starting with the left, the operator is applied to the next to operands:
So, (A+B)*C (operator needs to operands.
A+B is temp and considered an operand.
Of course, there are more advanced parse trees! (Different one in book)
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Post-Order Traversal – LRN.

*
You can guess the execution sequence:
Call itself to traverse the node’s left subtree
Call itself to traverse the node’s right subtree
Visit the node, N
A
 Will use the tree in the book.
 Priority: L R N in that order.
 So, ABC+*
Note: you will only pop after visiting a node…
+
B
My words: Start with the root.
Go left. Get A Can I go to the left. No. Can I go to the right? No.
 Therefore visit A
Go back to previous node (root here) Have gone left. Then go right. (I’m at +)
But before I visit +, any chance to go left. Yes! Go left. (I’m at B)
Any more to the left? No (I’m at B). Anything to the Right? No.
 So visit B.
Back to the node, +. Next priority is to the right or C.
But before I ‘visit’ C can I go to the left? NO. Can I go to the right? NO
 Ergo, Visit C
Back to the node, +. Have visited L and R. So N is left.
 Visit N, ( + ).
Now recurse to it’s parent. Have gone to the left; have gone to the right.
 So visit that node, *. We are done.
Priority: Go left first, then right, then the node (visit) L – R – N.
Note this is the postfix notation!!!!!
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This gives us: abc+* which is postfix
notation!!
C
Evaluating a Postfix Notation – postorder traversal…
 So, how would you evaluate an expression
in postfix notation?
 Create tree using postfix notation as input.
 Once tree is built, look for operators.
 Play with this…you will see these again.
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Enter: Pseudo Code for Iterative Scans:
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One GREAT way to approach these (as a few of you have emailed me) is to design the modules first; that is,
use pseudo-code to draft out your logic,
You all have progressed to the point where punching away at code is extremely frustrating and very error
prone. You can easily get to this point that you might mess up more than you fix. So it really is time to hone
your skills to the next level and try to design the logic before trying to implement your logic in code.
Please consider the following as an example: For inorder scan (LNR)
Inorder Scan: (LNR)
Clear stack // since we use the stack in three separate iterative scans, clear it prior to each one.
Set current to root node
// need this to get me started into the tree.
Loop while true
if current is not null
{
Push current onto stack
Set current to left-child of current // remember: “L” is top dog. Go to the left as
// much as possible
} // end if
//
before ‘visiting’ the node.
else
{
If stack is empty
// nothing left to pop; boundary condition.
return
Pop stack and set current equal to object popped from stack
Visit this node
// in this app, print it out
Set current to right child of current
} // end else
End loop
End inorder (LNR) scan.
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Pre-Order (NLR) Iterative Scan
Create a stack (or use current one, but clear it) // Stack myStack = new Stack();
// of course, Stack must have a push() pop(), isEmpty() isFull() etc. and
// these routines must adjust the stack index as expected…
Could pass the root as a parameter and pass it, as in
// NLR-Iterative-Scan (Node current) or something like this…
Pre-Order NLR Iterative Scan:
If current is not null // start at root. Make root current node.
{
Create a node for the Stack
Push onto Stack
// push node onto stack
Loop while stack is not empty
{
Pop Stack
// maybe current = (Node) myStack.pop()
Visit popped object // display the node in our application
If current.right child is not null
// this sequence is critical.
Push this object onto the stack
// myStack.push(the object)
If current.left child is not null
// note right child pushed before left
child.
// Why?
Push this object onto the stack
} // end loop
} // end if
} // end NLR iterative preorder scan
35 as obvious…)
Will leave iterative LRN scan for you…. (not
Finding Maximum and Minimum Values
 Pretty easy to do in a binary search tree.
 Do the LNR scan. When you encounter an
L with no left subtree, ‘that’ is the minimum.
 Maximum? Go to the right. Same song.
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