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Trees

- Terminology and Characterizations
- Spanning Trees
- Minimal Spanning Trees
- Binary Trees
- Tree Traversals
- Decision Trees
- Isomorphisms of Trees

Introduction

- Family trees use vertices to represent the

members of a family and edges to represent

parent-child relationships

Introduction

- A computer file system

Introduction

- A file system in Visual Studio .NET

Introduction

- A tree T is a simple graph satisfying If v and w

are vertices in T, there is a unique simple path

from v to w. - A rooted tree is a tree in which a particular

vertex is designated the root.

Introduction

- Which of the graphs are trees?

Terminology of Trees

- Let v0 be root, suppose that x, y and z are

vertices and that (v0, v1, ..., vn) is a simple

path. - vn-1 is the parent of vn.
- v0, v1, ..., vn-1 are ancestors of vn.
- vn is a child of vn-1.
- If x is an ancestor of y, y is descendant of x.
- If x and y are children of z, x and y are

siblings - If x has no children, x is terminal vertex (or

leaf) - If x is not terminal vertex, x is an internal

vertex.

Terminology of Trees

Properties of Trees

- Let T be a graph with n vertices. The following

are equivalent. - T is a tree.
- T is connected and acyclic.
- T is connected and has n-1 edges.
- T is acyclic and has n-1 edges.
- Prove if (d) then (a)

Binary Tree

- In a binary tree, an internal vertex has two

children, left child and right child. - The tree rooted at the left child of a vertex is

called the left subtree of this vertex.

Binary Tree

- A full binary tree is a binary tree in which each

vertex has either two children or no children. - If T is a full binary tree with i internal

vertices, then T has i1 terminal vertices and

2i1 total vertices. - If a binary tree of height h has t terminal

vertices, then lg t ? h

Huffman Coding

- When us bit strings to encode the letters of the

English alphabet. - We use 5 bits to represent each of 26 letters
- The total number of bits used to encode data is

five times the number of characters in the text. - Saving memory space, transmittal time.
- Is there a coding scheme to encode text with

fewer bits? - Huffman coding. (a graduate term paper by Huffman

at MIT) - The scheme assumes that we already know how many

times each letter occurs in the text!

Huffman Coding

- procedure Huffman (C symbols ai with frequencies

wi) - F forest of n rooted trees, each

consisting of - the single vertex ai and assigned

weight wi - while F is not a tree
- begin
- Replace the rooted trees T and T of

least weights - from F with w(T) w(T) with a tree

having a new - root that has T s its left subtree

and T as its right - subtree. Lable the new edge to T

with 0 and the - new edge to T with 1.
- Assign w(T) w(T) as the weight of

the new tree - end
- end Huffman

Huffman Coding

- Use Huffman coding to encode the following

symbols with frequencies listed A 0.08, B0.10,

C0.12, D0.15, E0.20, F0.35

Huffman Coding

- Use Huffman coding to encode these symbols with

given frequencies a 0.20, b 0.10, c 0.15, d

0.25, e 0.30. What is the average number of bits

required to encode a character?

Binary Search Trees

- A binary search tree is a binary tree in which
- each child of a vertex is designated as a right

or left child, with each vertex labeled with a

key - the key at a vertex is larger than the keys of

all vertices in its left subtree and smaller than

the keys of all vertices in its right subtree.

Binary Search Trees

- To build a binary search tree
- Create a root node, and assign the first item in

the list as the key of the root node - To add a new item, repeatedly compare the new

item with the keys of vertices already in the

tree - if the item is less than the key of the vertex,

move to the left - if the item is greater than the key of the

vertex, move to the right - If the item is less than the respective vertex

and this vertex has no left child, add a new

vertex with this item as a new left child. - If the item is greater than the respective vertex

and this vertex has no right child, add a new

vertex with this item as a new right child. - Repeat steps 2-4 add another items in the list to

the tree.

Binary Search Trees

- To build a binary search tree for words

mathematics, physics, gegraphy, zoology,

meteorology, geology, psychology, and chemistry

procedure make_bin_search_tree(w, n) let T

be the tree with one vertex, root store w1

in root for i 2 to n do begin

v root search true while

search do begin s

word in v if wi lt s then

if v has no left child then add a left

child l to v store wi in l

search false else v left child of

v else if v has

no right child then add a right child r to v

store wi in r search

false else v right child of

v end end end

make_bin_search_tree

Binary Search Trees

- To build a binary search tree for words
- Old Programmers Never Die They Just Lose Their

Memories

Decision Trees

- A decision tree is a rooted tree in which each

internal vertex corresponds to a decision, with a

subtree at these vertices for each possible

outcome of the decision.

Decision Trees

- To sort three numbers, a1, a2, and a3.

Tree Traversal

- Ordered rooted trees are used to store

information, such as arithmetic expressions

involving numbers, variables, and operations. - To traverse a tree, we use a labeling scheme

called universal address system of the ordered

rooted tree. - Label the root with the integer 0. Then label its

k children from left to right with 1, 2, 3, ...,

k. - For each vertex v at level n with label A, label

its children from left to right with A.1, A.2,

..., A.m.

Tree Traversal

Preorder Tree Traversal

- procedure preorder(PT)
- if PT is empty then
- return
- process PT
- preorder( the left subtree of PT )
- preorder( the right subtree of PT )
- end preorder

Preorder Tree Traversal

- In which order does a preorder traversal visit

the vertices in the ordered rooted tree T shown?

Inorder Tree Traversal

- procedure inorder(PT)
- if PT is empty then
- return
- inorder( the left subtree of PT )
- process PT
- inorder( the right subtree of PT )
- end inorder

Inorder Tree Traversal

- In which order does an inorder traversal visit

the vertices in the ordered rooted tree T shown?

Infix Form of an Expression

- Using inorder traversal, we get

Operands A, B, C, D, E Operators , -, , /

Arithmetic Expressions

- The expression ((xy)?2)((x-4)/3) may be

represented as a rooted binary tree.

Arithmetic Expressions

- What is the inorder traversal of the binary tree?
- infix form
- What is the preorder traversal?
- prefix form

Spanning Trees

- Consider the road system in Minnesota. The

highway department needs to plow the fewest roads

in winter, so that, there are always cleared

roads connecting any two towns. - But, how?
- To find a connected subgraph with the minimum

number of edges containing all vertices

Spanning Trees

Spanning Trees

- A spanning tree of G is a subgraph of G that is a

tree containing every vertex of G. - Example IP multicasting
- A computer sends a single copy of data over the

network. For data to reach receiving computers as

quickly as possible, there should be no loops.

Depth-First Search

- procedure dfs(V, E)
- w v1 V v1 E ??
- while V ? V do
- while there is edge (w, vk) with vk ? V
- add (w, vk) to E
- add vk to V
- w vk
- w parent of w in T //backtracking
- end dfs

Depth-First Search

- Choose a vertex as a root.
- Successively add vertices and edges, where each

new edge is incident with the last vertex in the

path and the new vertex is not in the path yet. - If the path goes through all vertices of the

graph, we have the spanning tree. - Otherwise, move back to the next to last vertex

in the path, and form a new path starting at this

vertex through vertices that were not visited yet.

Depth-First Search

- Find the spanning tree rooted at f

Breath-First Search

- Choose a root vertex
- Add all edges incident to this vertex.
- Order all newly added vertices
- For each new vertex, add each edge incident to

this vertex as long as it does not form a cycle - Repeat steps 3 and 4

Breath-First Search

- procedure bfs(V, E)
- S (v1), V v1, E ??
- while true do
- for each x?? S, in order, do
- if (x, y) is an edge not forming

a cycle then - add edge (x, y) to E
- add y to V
- if no edges were added then
- return(T)
- S children of S
- end bfs

Breath-First Search

- Find the spanning tree rooted at e

Minimum spanning Trees

- A minimum spanning tree of G is a spanning tree

of G with minimum weight. - Example a company plans to build a communication

network connecting its five computer enters. Any

pair of these centers can be linked with a leased

telephone line. Which links should be made to

ensure that there is a path between any two

computer centers so that the total cost of the

network is minimized?

Minimum spanning Trees

- The Prims Algorithm builds a tree by iteratively

adding edges until a minimal spanning tree is

obtained. At each iteration, it adds a

minimum-weight edge that does not complete a

cycle to the current tree. - Prims Algorithm is a greedy algorithm.

Minimum spanning Trees

- procedure prim(w, n, s)
- for i 1 to n do
- v(i) 0 v(s) 1 E ??
- for i 1 to n-1 do
- min ?
- for j 1 to n do
- if v(j) 1 then
- for k1 to n do
- if v(k) 0

and w(j, k) lt min then - add_vertex k
- e

(j, k) min w(j, k) - v( add_vertex ) 1
- E E ? e
- end prim

Prims Algorithm

Prims Algorithm

- Use Prims algorithm to find a minimum spanning

tree

Isomorphic Trees

- Let T1 and T2 be rooted trees with roots r1 and

r2. T1 and T2 are isomorphic if there is a

one-to-one, onto function f from the vertex set

of T1 to the vertex set of T2 satisfying - Vertices vi and vj are adjacent in T1 if and only

if the vertices f(vi) and f(vj) are adjacent in

T2. - f(r1) r2.

Nonisomorphisms Trees

- Three nonisomorphic trees with five vertices.

- Four nonisomorphic trees with four vertices.

Isomorphic Binary Trees

- Let T1 and T2 be binary trees with roots r1 and

r2. The T1 and T2 are isomorphic if there is a

non-to-one, onto function f satisfying - T1 and T2 are isomorphic tree.
- v is a left child of w in T1 iff f(v) is the left

child of f(w) in T2. - v is a right child of w in T1 iff f(v) is the

right child of f(w) in T2.

Isomorphic Binary Trees

- Five nonisomorphic binary trees with three

vertices

v1

Isomorphic Binary Trees

- procedure bin_tree_isom (r1, r2)
- if r1 null and r2 null then
- return (true)
- if r1 null or r2 null then
- return (false)
- lc_r1 left child of r1
- lc_r2 left child of r2
- rc_r1 right child of r1
- rc_r2 right child of r2
- return ( bin_tree_isom(lc_r1, lc_r2)
- and bin_tree_isom(rc_r1,

lc_r2) ) - end bin_tree_isom

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