trees

Data Structures and Algorithms 1

Trees

The definitions for this presentation are from from: Corman , et. al., Introduction to Algorithms (MIT Press), Chapter 5.Some material on binomial trees is from Hull.


A Few Applications

Arithmetic Expressions

b + a * b+

b *

a b


A Few Applications

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<name> <ssn> title

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XML Document Object Model

From Hull 4

A Few Applications

Binomial “Trees” Movements in Time t

Su

Sd

S

p

1 – p

Binomial trees are frequently used to approximate the movements in the price of a stock or other assetIn each small interval of time the stock price is assumed to move up by a proportional amount u or to move down by a proportional amount d.


Definitions

A Free tree is a connected, acyclic undirected graph.


If an undirected graph is acyclic but possibly disconnected,it is a forest.


This is a graph that contains a cycle and is therefore neither atree nor a forest.


Theorem (Properties of Free Trees)

Let G = (V, E) be an undirected graph.The following statements are equivalent:

1. G is a free tree.


2. Any two vertices in G are connected by a unique simple path.


3. G is connected, but if any edge is removed from E, the resulting graph is disconnected.


4. G is connected, and |E| = |V| - 1.


5. G is acyclic, and |E| = |V| - 1.


6. G is acyclic, but if any edge is added to E, the resulting graph contains a cycle.


A rooted tree is a free tree in which one of the vertices isdistinguished from the others. The distinguished vertex is called the root of the tree. We often refer to a vertex of arooted tree as a node (we may also call this a vertex) of thetree. The following figure shows a rooted tree on a set of 12nodes with root 7.

7

3 10 4

8 12 211

9

6 5 1


Consider a node x in a rooted tree T with root r. Anynode y on the unique path from r to x is called an ancestorof x. If y is an ancestor of x, then x is a descendant of y.(Every node is both an ancestor and a descendant of itself.)If y is an ancestor of x and x y, then y is a proper ancestorof x and x is a proper descendant of y. The subtree rootedat x is the tree induced by descendants of x, rooted at x.

8

6 5

9

r

x

y


If the last edge on the path from the root r of a tree T to anode x is (y, x), then y is the parent of x, and x is a child ofy. The root is the only node in T with no parent. If twonodes have the same parent, they are siblings. A node withno children is an external node or leaf. A nonleaf node isan internal node.


The number of children of a node x in a rooted tree T is called the degree of x. The length of the path from the root r to a node x is the depth of x in T. The largest depth of any node in T is the height of T.

7

3 10 4

8 12 211

9

6 5 1

depth 0

depth 1

depth 2

depth 3

depth 4

height = 4


An ordered tree is a rooted tree in which the children of each node are ordered. That is, if a node has k children,then there is a first child, a second child, …, and a kthchild. The two trees shown below are different whenconsidered to be ordered trees, but the same when considered to be just rooted trees.

7

3 10 4

8 12 211

9

6 5 1

7

561

211812

4103

9


A binary tree T is a structure defined on a finite set ofnodes that either

• contains no nodes, or• is comprised of three disjoint sets of nodes:

a root node, a binary tree called its left subtreeand a binary tree called its right subtree.

32 7

4 1 5

6


Full binary tree: each node is either a leaf or has degreeexactly 2.

In a positional tree, the children of a node are labeled with distinct positive integers. The ith child of a node isabsent if no child is labeled with integer i. A k-ary treeis a positional tree in which for every node, all childrenwith labels greater than k are missing. Thus, a binary tree is a k-ary tree with k = 2.


A complete k-ary tree is a k-ary tree in which all leaveshave the same depth and all internal nodes have degreek.

height = 3

depth 0

depth 1

depth 2

depth 3


How many leaves L does a complete binary tree of height hhave?

d = 0

d = 1

d = 2

The number of leaves at depth d = 2d

If the height of the tree is h it has 2h leaves.

L = 2h.


What is the height h of a complete binary tree with L leaves?

leaves = 1 height = 0



leaves = L height = Log2L

Since L = 2h

Log2L = Log22h

h = Log2L


The number of internal nodes of a complete binary tree of height h is ?

Internal nodes = 0 height = 0

Internal nodes = 1 height = 1

Internal nodes = 1 + 2 height = 2

Internal nodes = 1 + 2 + 4 height = 3

1 + 2 + 22 + . . . + 2 h-1 = 2i

= 2h - 1 2 - 1

Thus, a complete binary tree of height = h has 2h-1 internalnodes.

Geometric series


The number of nodes n of a complete binarytree of height h is ?

nodes = 1 height = 0



nodes = 2h+1- 1 height = h

Since L = 2h

and since the number of internal nodes = 2h-1 the total number of nodes n = 2h+ 2h-1 = 2(2h) – 1 = 2h+1- 1.


If the number of nodes is n then what is the height?




nodes = n height = Log2(n+1) - 1

Since n = 2h+1-1n + 1 = 2h+1

Log2(n+1) = Log2 2h+1

Log2(n+1) = h+1h = Log2(n+1) - 1


n

n

n

2

1

1= 1 (2n)! (n+1) n! (2n-n)!

Catalan Numbers

1,1,2,5,14,...


The number of distinct binary trees with n nodes

n

n

n

2

1

1N = 1 N=2

...

N=3N = 0


Class for Binary Nodes

public class BTNode { private Object data; private BTNode left; private BTNode right; ...


public BTNode(Object obj, BTNode l, BTNode r) { data = obj; left = l; right= r; } public boolean isLeaf() { return (left == null) && (right == null); } ...


Copying Trees public static BTNode treeCopy(BTNode t) { if (t == null) return null; else { BTNode leftCopy = treeCopy(t.left); BTNode rightCopy = treeCopy(t.right); return new BTNode(t.data, leftCopy, rightCopy); } }


Tree Traversals

•Preorder•Inorder•Postorder•Levelorder


public void preOrderPrint(){System.out.println(data);if (left != null)

left.preOrderPrint();if (right != null)

right.preOrderPrint();}

a b c d e f g a

b e

c d f g

Root, Left, Right


public void inOrderPrint(){if (left != null) {

left.inOrderPrint()System.out.println(data);if (right != null)

right.inOrderPrint()}

c b d a f e g a

b e

c d f gLeft, Root, Right


public void postOrder(){if (left != null

left.postOrder()if (right != null)

right.postOrder()System.out.println(data);

}

c d b f g e a

a

b e

c d f g

Left, right, root


levelorder (T) { Q = makeEmptyQueue()

enqueue (T,Q)until isempty (Q) {

p = dequeue(Q)visit (p)for each child of P, in order, do

enqueue (, Q)}

}

a b e c d f g

a

b e

c d f g


An Array Representation

Suppose we are lucky enough to be working with complete binary trees.

We can store the tree in an array.

Let the root be at index 0 and let the left and right childrenof node i be at indexes 2i+1 and 2i+2 respectively.

Lewis and Denemberg, Page 111


An Array Representation

isLeaf(i) : 2i + 1 >= nleftChild(i) : 2i + 1 (none if 2i+ 1 >= n)rightChild(i): 2i + 2 (none if 2i + 2 >= n)leftSibling(i): i - 1 (none if i == 0 or i is odd)rightSibling(i) : i + 1 (none if i = n-1 or i is even) parent(i) = int((i-1)/2) (none if i == 0)

Lewis and Denemberg, Page 111

Works if the tree is “almost complete”, growing top to bottomand left to right.


Binomial “Trees” Using an Array Representation

S0

S0u

S0d S0 S0

S0u 2

S0d 2

S0u 2

S0u 3 S0u 4

S0d 2

S0u

S0d

S0d 4

S0d 3 The array element bt[0] will be S0.In general, given a node with index i at depth d, its left child is located at bt[i + d+1] and its right child is located at bt[i + d+2]

trees

Documents

depth of x

ancestor of x

parent of x

anode x

descendants of x

abdata structures

rxydata structures

rooted tree t