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Data Structures Review Session 2

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Data Structures Review Session 2

Ramakrishna, PhD student.

Grading Assistant for this course

- A binary tree is a tree where each node has at most two children, referred to as the left and right child
- A binary search tree is a binary tree where every node's left subtree holds values less than the node's value, and every right subtree holds values greater.
- A new node is added as a leaf.

root

parent

17

right child

11

19

left child

Problem 1:

How do you sort n numbers using a binary

Search tree ?? What are the best, worst and

Average case time complexities ??

- All dynamic-set operations (Search, Insert, Delete, Min, Max, successor, predecessor) can be supported in O(h) time.
- h = (lg n) for a balanced binary tree (and for an average tree built by adding nodes in random order.)
- h =(n) for an unbalanced tree that resembles a linear chain of n nodes in the worst case.
- Red-black trees are a variation of binary search trees to ensure that the tree is balanced.
- Height is O (log n), where n is the number of nodes.

- A binary tree can be balanced, such that one branch of the tree is about the same size and depth as the other.
- In order to find out if a tree node is balanced, you need to find out the maximum height level of both children in each node, and if they differ by more than one level, it is considered unbalanced.
- If the number is -1, 0, or 1, the node is balanced. If the difference is anything else, then it is unbalanced. Note that a balanced binary tree requires every node in the tree to have the balanced property.

Inorder traversal of a binary search tree always gives a sorted

sequence of the values. This is a direct consequence of the BST

property.

Given a set of unordered elements, the following method can be

used to Sort the elements:

- construct a binary search tree whose keys are those elements, and then
- perform an inorder traversal of this tree.
BSTSort(A)1. for each element in an array do 2. Insert element in the BST// Constructing a BST take O( log n) time 3. Inorder-Traversal (root) // Takes O(n) time

Best case running time of BSTSort(A) is O( n log n).

Worst case running time is O(n2) since each inserting could take O(n) time in worst case.

Input Sequence :- 2 3 8 4 5 7 6 1 2.5 2.4

Step 1 : Creating Binary Search Tree of above given input sequence.

2

3

1

8

2.5

4

2.4

5

7

6

Input Sequence :- 2 3 8 4 5 7 6 1 2.5 2.4

Step 2 : Perform Inorder-Traversal.

2

1

2

3

1

8

2.5

4

2.4

5

7

6

Input Sequence :- 2 3 8 4 5 7 6 1 2.5 2.4

Step 2 : Perform Inorder-Traversal.

2

1

2

3

1

2.4

2.5

3

8

2.5

4

2.4

5

7

6

Input Sequence :- 2 3 8 4 5 7 6 1 2.5 2.4

Step 2 : Perform Inorder-Traversal.

2

1

2

3

1

2.4

2.5

3

8

2.5

4

5

6

4

2.4

7

8

5

Sorted Array

7

6

Input Sequence :- 8 4 3 2 and 2 3 8 10

Step 1 : Creating Binary Search Tree of above given input

sequence.

2

8

4

3

3

8

2

10

Tree-Minimum(x)Tree-Maximum(x)

1. whileleft[x] NIL 1. whileright[x] NIL

2. dox left[x] 2. dox right[x]

3. returnx 3. returnx

- The binary-search-tree property guarantees that:
- The minimum is located at the left-most node.
- The maximum is located at the right-most node.

Q: How long do they take?

int vdpth v.depth

int wdpth w.depth

while vdpth > wdpth do

v v.parent

vdpth vdpth -1

end while

while wdpth > vdpth do

w w.parent

wdpth wdpth -1

end while

while v ≠ w do

v v.parent

w w.parent

end while

return v

Note that LCA algorithm is applicable for any tree

- Give an efficient algorithm for converting an infix arithmetic expression into its equivalent postfix notation ?
(Hint: First convert the infix expression into its equivalent binary tree representation)

Input: Fully-parenthesized arithmetic Expression

Output: A binary Tree T representing Expression

S a new empty stack

For i 0 to n-1 do {

if E[i] is a variable or an operator then

T new binary tree with E[i] as root

S.push(T)

else if E[i] = “(“ then continue;

else if E[i] = “)”

T2 S.pop()

T S.pop()

T1 S.pop()

Attach T1 as T’s left subtree and T2 as its right subtree

S.push(T)

}

Return S.pop()

Input: Full-Parenthesized arithmetic Expression ,E.

Output: PostFix notation of E

T buildExpressionTree (E)

Postorder-Traversal (T.root) // this gives the

Postfix notation of the expression

Preorder-Traversal (T.root) // this gives the

Prefix notation of the expression

Inorder-Traversal (T.root) // this gives the

Infix notation of the expression

- Assume V = {1, 2, …, n}
- An adjacency matrixrepresents the graph as a n x n matrix A:
- A[i, j] = 1 if edge (i, j) E (or weight of edge)= 0 if edge (i, j) E

- Example:

1

a

d

2

4

b

c

3

How much storage does the adjacency matrix require?

A: O(V2)

- Adjacency list: for each vertex v V, store a list of vertices adjacent to v
- Example:
- Adj[1] = {2,3}
- Adj[2] = {3}
- Adj[3] = {}
- Adj[4] = {3}

- Variation: can also keep a list of edges coming into vertex

1

2

4

3

- Adjacency lists take O(V+E) storage

- Given an adjacency-list representation of a directed graph, how long does it take to compute the out-degree and in-degree of every vertex ? Also, how long is it going to take if we use the adjacency matrix instead ?

- Given the adjacency-Matrix representation of a graph the out and in-degree of every node can easily be computed as follows.
For I 1 to v // v vertices

For J 1 to v // v vertices

if(A[I][J] == 1) then

outdegree[I]++;

indegree[J]++;

Time complexity : O(V2)

- Given the adjacency-List representation of a graph the out and in-degree of every node can easily be computed as follows.
For I 1 to v // n vertices

For J 1 to A[I].size() // neighbours of vertex I

n A[i].get(J)

outdegree[I]++;

indegree[n]++;

Time complexity : O(V+E)

- The transpose of a directed graph G = (V,E) is the graph G with all its edges reversed. Describe efficient algorithms for computing the transpose of G for both adjacency list and matrix representations. What are the running times ?

- Given the adjacency-Matrix representation of a graph the transpose can be computed as follows.
For I 1 to v // v vertices

For J 1 to v // v vertices

Transpose[I][J] = A[J][I]

Time complexity : O(V2)

- Given the adjacency-List representation of a graph the transpose can be computed as follows.
For I 1 to n // n vertices

For J 1 to A[I].size() // neighbours of vertex I

Transpose[I].add(A[J].get(I))

Time complexity : O(V+E)

- Successor of node x is the node y such that key[y] is the smallest key greater than key[x].
- The successor of the largest key is NIL.
- Search consists of two cases.
- If node x has a non-empty right subtree, then x’s successor is the minimum in the right subtree of x.
- If node x has an empty right subtree, then:
- As long as we move to the left up the tree (move up through right children), we are visiting smaller keys.
- x’s successor y is the node that x is the predecessor of (x is the maximum in y’s left subtree).
- In other words, x’s successor y, is the lowest ancestor of x whose left child is also an ancestor of x.

26

200

28

190

213

18

56

12

24

27

Tree-Successor(x)

- if right[x] NIL
2. then return Tree-Minimum(right[x])

3. yp[x]

4. whiley NIL and x = right[y]

5. dox y

6. yp[y]

7. returny

Code for predecessor is symmetric.

Running time:O(h)

Note that external nodes are also called leaf nodes

Let

# leaf nodes = E, # internal nodes = I , Height = H, # nodes = N,

- E = I + 1
- N = E + I
- (# nodes at level x) <= 2x
- E <= 2H
- H >= log2 I
- A full binary tree has (2H+1 – 1) nodes.
- (H+1) ≤ E ≤ 2H.
- H≤ I ≤ 2H -1.
- 2H+1 ≤ N ≤ 2H -1.
- log2(N+1)-1 ≤ H ≤ (N-1)/2
- log2(E) ≤ H ≤ E-1
- log2(I+1) ≤ H ≤ I