1. Data Structures Review Session 2
Ramakrishna, PhD student.
Grading Assistant for this course
CS 307 Fundamentals of Computer Science

2. Binary Search Trees
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 11 19
right child
left child
CS 307 Fundamentals of Computer Science

3. Problems Problem 1: How do you sort n numbers using a binary
Search tree ?? What are the best, worst and
Average case time complexities ??
CS 307 Fundamentals of Computer Science

4. Binary Search Tree properties
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.
CS 307 Fundamentals of Computer Science

5. Balance of Binary Trees
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.
CS 307 Fundamentals of Computer Science

6. Sorting Using BST
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.
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7. Example Sorting Using BST
Input Sequence :
Step 1 : Creating Binary Search Tree of above given input sequence. 2 3 1 8
2.5 4
2.4 5 7
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8. Example Sorting Using BST (cont.)
Input Sequence :
Step 2 : Perform Inorder-Traversal. 2 1 2 3 1 8
2.5 4
2.4 5 7 6
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9. Example Sorting Using BST (cont.)
Input Sequence :
Step 2 : Perform Inorder-Traversal. 2 1 2 3 1
2.4
2.5 3 8
2.5 4
2.4 5 7 6
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10. Example Sorting Using BST (cont.)
Input Sequence :
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
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11. Worst cases 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
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12. Finding Min & Max Tree-Minimum(x) Tree-Maximum(x)
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.
Tree-Minimum(x) Tree-Maximum(x)
1. while left[x]  NIL while right[x]  NIL
do x  left[x] do x  right[x]
3. return x return x
Q: How long do they take?
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13. Algorithm LCA (Node v, Node w):
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
while v ≠ w do
return v
Note that LCA algorithm is applicable for any tree
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14. 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)
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15. CS 307 Fundamentals of Computer Science

16. Algorithm buildExpressionTree (E):
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 }
Return S.pop()
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17. Algorithm PostFix (E):
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
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18. Representing Graphs Assume V = {1, 2, …, n}
An adjacency matrix represents 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
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19. Graphs: Adjacency Matrix
Example: A 1 2 3 4 1 a 2 d 4 b c 3
How much storage does the adjacency matrix require?
A: O(V2)
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20. Graphs: Adjacency List
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
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21. Graph Problems
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 ?
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22. Solution
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)
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23. Solution
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)
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24. Problem
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 ?
CS 307 Fundamentals of Computer Science

25. Solution
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)
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26. Solution
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)
CS 307 Fundamentals of Computer Science

27. Predecessor and Successor in a BST
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.
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28. Pseudo-code for Successor
Tree-Successor(x)
if right[x]  NIL
then return Tree-Minimum(right[x])
y  p[x]
while y  NIL and x = right[y]
do x  y
y  p[y]
return y 56 26
200 18 28
190
213 12 24 27
Code for predecessor is symmetric.
Running time: O(h)
CS 307 Fundamentals of Computer Science

29. Properties of Binary Trees
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
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