Binary Trees 2 Prof. Sin-Min Lee Department of Computer Science.

Binary Trees 2

Prof. Sin-Min Lee

Department of Computer Science

For each number of nodes, n, there is a certain number of possible binary tree configurations. These numbers form a sequence of integers with respect to n.

A useful way to describe an integer sequence is to construct a generating

function:

whose coefficients bi are the sequence.

B(x) is power series over a purely formal variable x.

Apparently, b1 is 1, and b2 is 2. The b0 coefficient

is somewhat artificial, its the no-nodes tree which is, I guess, the only one. Further analysis gives the following idea: if a binary tree has n nodes, then there must be one node as the root and two subtrees. The total number of nodes in the subtrees must be n-1, and either of them may be empty. Assuming k nodes in the left subtree, the right subtree then has n-k-1 nodes, k going from 0 to n-1. The root node is always there, and therefore the tree configurations differ only by the configuration of subtrees.

The total number of possible trees on n nodes can then be expressed as:

Postorder

Template <class T>void Postorder ( BinaryTreeNode<T> *t){ // Postorder traversal of *t. If (t){ Postorder ( t->LeftChild);//do left subtree Postorder(t->RightChild);//do right subtree visit(t); // visit tree root }}Each root is visited after its left and right subtrees have been traversed.

Example

Preorder +*ab/cdInorder a*b+c/dPostorder ab*cd/+

Elements of a binary tree listed in pre-,in-,and postorder.

+

b

* /

dca

About level order

While loop only if the tree is not emptyFollowing the addition of the children of t to the queue,we attempt to delete an element from the queue. If the queue is empty,Delete throws an OutOfBounds exception. If the queue is not empty,then Delete returns the deleted element in t.

Sample tree to illustrate tree traversal

1

2 3

5

10

4

8 9

6

11 12

7

Tree after four nodes visited in preorder traversal

1

2 3

5

10

4

8 9

6

11 12

7

1

2

3

4

Tree after left subtree of root visited using preorder traversal

1

2 3

5

10

4

8 9

6

11 12

7

1

2

3

4

6

5 7

Tree after completed preorder traversal

1

2 3

5

10

4

8 9

6

11 12

7

1

2

3

4

6

5 7

8

9

10 11

12

Tree visited using inorder traversal

1

2 3

5

10

4

8 9

6

11 12

7

7

4

2

1 3

5

6

11

9

8 10

12

Tree visited using postorder traversal

1

2 3

5

10

4

8 9

6

11 12

7

12

6

3

1 2

5

4

11

9

7 8

10

Here we will work through an example showing how a preorder traversal can be obtained by explicit use of a stack. The logic

follows (see the text for the code):

1.Stack the root node. 2.Exit with completion if the stack is

empty. 3.Pop the stack and visit the node

obtained. 4.Push the right child, if it exists, on

the stack. 5.Push the left child, if it exists, on

the stack. 6.Go to 2.

Using the runtime stack via recursion, we can write simple definitions of the three depth-first traversals without use of an

explicitstack (in these definitions I leave out the templates for

brevity):

void Preorder(TreeNode *p) { if (p) {

Visit(p); Preorder(p->left);

Preorder(p->right); }}

void Inorder(TreeNode *p) { if (p) {

Inorder(p->left); Visit(p);

Inorder(p->right); }}

void Postorder(TreeNode *p) { if (p) {

Postorder(p->left); Postorder(p->right);

Visit(p); }}

TREE SORTCombines tree initialization, insertion, and inorder traversal to

generate and print each node in a "sorted" binary tree.1. Initialize: read first item, create root node.

2. For each item after the first: read item and insert in tree.3. Traverse in order, printing each node.

More formally, in pseudocode:def treesort ()

read item root = makebt(null, item, null)

while not end of data read item

insert(root, item) end while

inorder(root)end

Using tree sort, sort the letters M, A, R, J, K, S, V.

Insertion produces M / \

A R \ \

J S \ \

K V

Traverse in order => A J K M R S V