Trees

1
Trees

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

2
Introduction

• Family trees use vertices to represent the
  members of a family and edges to represent
  parent-child relationships

3
Introduction

• A computer file system

4
Introduction

• A file system in Visual Studio .NET

5
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.

6
Introduction

• Which of the graphs are trees?

7
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.

8
Terminology of Trees
9
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)

10
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.

11
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

12
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!

13
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

14
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

15
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?

16
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.

17
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.

18
Binary Search Trees

• To build a binary search tree for words
  mathematics, physics, gegraphy, zoology,
  meteorology, geology, psychology, and chemistry

19
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
20
Binary Search Trees

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

21
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.

22
Decision Trees

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

23
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.

24
Tree Traversal
25
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

26
Preorder Tree Traversal

• In which order does a preorder traversal visit
  the vertices in the ordered rooted tree T shown?

27
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

28
Inorder Tree Traversal

• In which order does an inorder traversal visit
  the vertices in the ordered rooted tree T shown?

29
Infix Form of an Expression

• Using inorder traversal, we get

Operands A, B, C, D, E Operators , -, , /
30
Arithmetic Expressions

• The expression ((xy)?2)((x-4)/3) may be
  represented as a rooted binary tree.

31
Arithmetic Expressions

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

32
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

33
Spanning Trees
34
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.

35
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

36
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.

37
Depth-First Search

• Find the spanning tree rooted at f

38
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

39
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

40
Breath-First Search

• Find the spanning tree rooted at e

41
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?

42
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.

43
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

44
Prims Algorithm
45
Prims Algorithm

• Use Prims algorithm to find a minimum spanning
  tree

46
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.

47
Nonisomorphisms Trees

• Three nonisomorphic trees with five vertices.

• Four nonisomorphic trees with four vertices.

48
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.

49
Isomorphic Binary Trees

• Five nonisomorphic binary trees with three
  vertices

v1
50
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