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Chapter 2:-
Analysis of Algorithm
Compiled By:- Sanjay Patel
Assistant Professor,
SVBIT.
Outline
The efficient algorithm
Average and worst case analysis
Elementary operation
Asymptotic notation
Analyzing control statement
Amortized analysis
Sorting algorithm
Binary tree search
Sanjay Patel
Empirical approach(Posteriori approach)
Theoretical approach(Priori)
Hybrid approach
The efficient algorithm
Sanjay Patel
We write the programs for various competing
techniques and then that program is tested on the
computer using different instances.
This approach is also called as posteriori approach.
Disadvantages
Practical consideration may force us to test
algorithm for small number of instances.
Discovering the algorithm that performs better for
large as well as for small instances is critical.
Empirical approach(Posteriori approach)
Sanjay Patel
The requirement quantity of resources is determined
mathematically. The resources that are considered in
this approach are-computing time and storage
space.
This approach is also known as priori approach.
This approach is independent of the computer being
used, programming language used and skill of the
programmer.
It saves time spent in programming an inefficient
algorithm and testing them.
Theoretical approach(Priori)
Sanjay Patel
In this approach the function that describes the
algorithm’s efficiency is determined theoritically and
then any required numerical parameters are
determined empirically when program executes on
some machine.
This approach is called as an hybrid approach
because it makes use of theoretical and empirical
approaches in combination.
The time required by an algorithm for very large
instances can be computed , but the predictions must
be made with the help of theoritical approach.
Hybrid approach
Sanjay Patel
Functions in order of increasing growth rate
Sanjay Patel
Function Name
C Constant
LogN Logarithmic
Log2N Log-squared
N Linear
NlogN NlogN
N2 Quaratic
N3 Cubic
2n Exponential
Common Computing time function
Sanjay Patel
Arrange following rate of growth in increasing
order.
2n,NlogN, N2, N ,1, LogN ,N!, N3
GTU Dec-2010 (2 marks)
Sanjay Patel
Predict resource utilization
1. Memory (space complexity)
2. Running time (time complexity)
Resource Utilization
Sanjay Patel
Best Case time complexity:- when algorithm
runs for short time.
Worst Case time complexity:- when algorithm
runs for a longest time.
Average Case time complexity:- Gives
information about the behavior of an algorithm
on specific or random input.
Three cases of analysis
Sanjay Patel
Elementary operation are those operation whose
execution time is bounded by a constant which
depends upon the type of implementation used.
The elementary operations are addition,
multiplication and assignment.
Elementary operation
Sanjay Patel
Algorithm SUM(n)
{
Sum 0
for (i1 to n)
do
sum sum + I
return sum;
}
In above algorithm the elementary operation is
Addition.
Example..
Sanjay Patel
Asymptotic notation
Sanjay Patel
To choose the best algorithm, we need to check efficiency of each algorithm.
The efficiency can be measured by computing time complexity of each algorithm.
Asymptotic notation is a shorthand way to represent the time complexity.
Classifying Functions by Their Asymptotic Growth
Theta (Θ)
Big Oh (O)
Big Omega (Ω)
Its denote by “O”
Definition: Let f(n) and g(n) be two functions; We
say that
f(n) O(g(n))
if there exists a constant c, no > 0 such that
f(n)≤c*g(n) for all n ≥ Xo.
f (n) is asymptotically less than or equal to g(n)
Big-O gives an asymptotic upper bound
BIG OH notation
Sanjay Patel
Consider function
F(n)= 2n + 2
g(n)= n2
initially, start with n=1, then n=2,3,4……
Thus Upper bound of exiting time is obtained by Big
OH notation.
EXAMPLE
Sanjay Patel
It is denoted by “ ”.
Definition: Let f (n)and g(n) be two functions; We
say that
f(n) (g(n))
if there exists a constant c ≥ 0 such that
f(n) ≥ c*g(n) for all n ≥ no
f(n) is asymptotically greater than or equal to g(n)
Big-Omega gives an asymptotic lower bound
OMEGA
Sanjay Patel
Consider example
F(n)= 2n2 + 5
g(n)=7n
Initially, start with n=0, then n=1,2,3,4……
EXAMPLE
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Definition: Let f(n) and g(n) be two functions ; We
say that
f(n) (g(n))
if there exists a constant c1, c2, no > 0; such that for
every integer n n0 we have
c1g(n) <= f(n) <= c2g(n)
F(n) is asymptotically equal to g(n)
F(n) is bounded above and below by g(n)
Big-Theta gives an asymptotic equivalence
THETA
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Consider example
f(n)= 2n+8
g(n)=7n
Where n>=2. ,,here c1=5, c2 = 7
EXAMPLE
Sanjay Patel
Use the informal definitions of “O”, “ ”and to
determine whether the following assertions are true
or false.
1) n(n+1)/2 € O(n3 )
2) n(n+1)/2 € O(n2 )
3) n(n+1)/2 € (n3 )
4) n(n+1)/2 € (n)
Example
Sanjay Patel
The recurrence equation is an equation that defines
a sequence recursively.
The recurrence relation can be solved by following
methods.
Substitution method
Master’s method
Recurrence Equation
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The substitution method is a kind of method in which
a guess for the solution is made.
There are two types of substitutions.
Forward substitution
Backward substitution
Substitution Method
Sanjay Patel
consider recurrence relation
T(n)= T(n-1) +n
with initial condition T(0)= 0, the n=1,2,3…
By observing each formula, we can derive a formula
T(n)= n(n+1)2
We can denote T(n) in terms of big oh notation
T(n)= O(n2)
Forward substitution
Sanjay Patel
consider recurrence relation
T(n)= T(n-1) +n
with initial condition T(0)= 0.
In backward substitution, start from n=n-1, n-2…
Let, n=n-1
Then T(n-1)= T(n-1-1) + n-1 …….(1)
Now , put (1) equation in original T(n) function
T(n) = T(n-2) + (n-1) +n ….
Backward substitution
Sanjay Patel
Now, n=n-2
T(n-2)= T(n-2-1)+(n-2)
=T(n-3) + (n-2)
Now, put the value of T(n-2) in previous equation.
So, T(n)= T(n-3) + (n-2) + (n-1) + n
::::::::::::::::::::::::::::::::::::::
= T(n-k) + (n-k+1) + (n-k+2)+…….+n
So if k=n then
T(n)= T(0) + 1 +2 +…….n
= 0+1 +2 + ………+n
T(n)= n(n+1)2
Cont’d…
Sanjay Patel
By observing each formula, we can derive a formula
T(n)= n(n+1)2
We can denote T(n) in terms of big oh notation
T(n) € O(n2)
Cont’d…
Sanjay Patel
Solve the x(n)= x(n-1) + 5 for n>1, x(1)=0
With recurrence relation.
Example
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Let solve this example with backward substitution.
n= n-1 .. X(n)= X(n-1) + 5
X(n-1)= X(n-1-1) +5
Put the value of X(n-1) into x(n)
so X(n)= [X(n-2)+5] +5
::::::::::::::::::::::::::
= [ X(n-3) + 5] +5 +5
:::::::::::::::::::::::::::::::
= X(n-i) + 5* i.
solution
Sanjay Patel
Now in example we have value X(1)=0
So if i=n-1 then we will get value of X(1)
X(n)= X(n-(n-1) + 5* (n-1)
= X(1) + 5 * (n-1)
= 0 + 5 *(n-1) X(1)=0
= 5 (n-1)
Cont’d….
Sanjay Patel
Solve the following recurrence relation
T(n)=T(n-1)+1 with T(0)=0 as initial condition . Find
the big oh notation.
Example
Sanjay Patel
It is efficiency analysis.
Consider the following recurrence relation
T(n)= a T(n/b) + F(n) where n>=d , d is constant
If F(n) is (nd ) where d>=0 in the recurrence relation
then.
Master Theorem
Sanjay Patel
T(n)= (nd) If a< bd
T(n)= (nd log n) If a=b
T(n)= (nlogba ) If a>bd
Find the growth for solution of the following
recurrence.
1. T(n)= 4T(n/2) + n T(1)=1
2. T(n)= 4T(n/2) + n2 T(1)=1
3. T(n)= 4T(n/2) + n3 T(1)=1
Example
Sanjay Patel
1. T(n)= 4T(n/2) + n T(1)=1
here, a=4, b=2 and F(n)= n1
so d=1
it is matching with case 3 because a>bd
4 > 2
i.e T(n)= (nlog24 )
But log24 = 2
T(n)= (n2) is the time complexity.
solution
Sanjay Patel
Sequencing
For loops
While loops
Recursive call
While and repeat loops
Analyzing control statement
Sanjay Patel
In amortized analysis means finding average
running time per operation over worst case
sequence of operation.
Aggregate Method
Accounting Method
Potential Method
Amortized analysis
Sanjay Patel
• Aggregate analysis is a kind of analysis in which the analysis is made over sequence of n operation and these operation actually take worst case time T(n).
• In the worst case, the average cost, or amortized cost , per operation is T(n)/n.
• The amortized cost applies to each operation, even when there are several types of operations in the sequence
Aggregate Method
Sanjay Patel
Stack example
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3 ops:
Push(S,x) Pop(S)Multi-
pop(S,k)
Worst-case
cost:O(1) O(1)
O(min(|S|,k)
= O(n)
The accounting method is based on the charges that are
assigned to each operation.
Assign different charges to different operation
The amount of charges is called amortized cost
Then there is actual cost of each operation.
The amortized cost can be more or less than actual cost.
When amortized cost > actual cost , the difference is saved in
specific object called credits.
When any particular operation amortized cost < actual cost
the stored credit are utilized.
Accounting method
Sanjay Patel
• Charge i th operation a fictitious amortized cost ĉi, where $1 pays for 1 unit of work (i.e., time).
• Assign different charges (amortized cost ) to different operations
• Some are charged more than actual cost
• Some are charged less
• This fee is consumed to perform the operation.
• Any amount not immediately consumed is stored in the bank for use by subsequent operations.
• The bank balance (the credit) must not go negative! We must ensure
Accounting method
Sanjay Patel
Stack Example
Sanjay Patel
3 ops:
Push(S,x) Pop(S) Multi-pop(S,k)
•Assigned
cost:2 0 0
•Actual cost: 1 1 min(|S|,k)
Push(S,x) pays for possible later pop of x.
1
1
1
1
1
1
1
1
1
0
1
1
0
0
0
• When pushing an object, pay $2
• $1 pays for the push
• $1 is prepayment for it being popped by either pop or Multipop
• Since each object has $1, which is credit, the credit can never go negative
• Therefore, total amortized cost = O(n), is an upper bound on total actual cost
Cont’d…
Sanjay Patel
This method is similar to the accounting method in
which the concept of “prepaid” is used.
Accounting method stores credit with specific objects
while potential method stores potential in the data
structure as a whole.
Can release potential to pay for future operations
Most flexible of the amortized analysis methods.
Potential method
Sanjay Patel
Cont’d..
Sanjay Patel
Framework:
• Start with an initial data structure D0. hence
for n operation D1 D2 D3..…• Operation i transforms Di–1 to Di.
• The cost of operation i is ci.
• Define a potential function : {Di} R,such that (D0 ) = 0 and (Di ) 0 for all i.
• The amortized cost ĉi with respect to is
defined to be ĉi = ci + (Di) – (Di–1).
Stack Example
Sanjay Patel
Define: (Di) = #items in stack Thus, (D0)=0.
Plug in for operations:
Push: ĉi = ci + (Di) - (Di-1)
= 1 + j - (j-1)
= 2
Pop: ĉi = ci + (Di) - (Di-1)
= 1 + (j-1) - j
= 0
Multi-pop: ĉi = ci + (Di) - (Di-1)
= k’ + (j-k’) - j k’=min(|S|,k)
= 0
Thus O(1) amortized time per op.
Bubble sort
Selection sort
Radix sort
Insertion sort
Shell sort
Heap sort
Sorting algorithm
Sanjay Patel
Bubble sort
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Sorting takes an unordered collection and makes
it an ordered one.
512354277 101
1 2 3 4 5 6
5 12 35 42 77 101
1 2 3 4 5 6
Cont’d..
Sanjay Patel
512354277 101
1 2 3 4 5 6
Swap42 77
Cont’d..
Sanjay Patel
512357742 101
1 2 3 4 5 6
Swap35 77
Cont’d..
Sanjay Patel
512773542 101
1 2 3 4 5 6
Swap12 77
Cont’d..
Sanjay Patel
577123542 101
1 2 3 4 5 6
No need to swap
Cont’d..
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577123542 101
1 2 3 4 5 6
Swap5 101
Cont’d..
Sanjay Patel
77123542 5
1 2 3 4 5 6
101
Largest value correctly placed
Bubbling all element
Sanjay Patel
77123542 5
1 2 3 4 5 6
101
5421235 77
1 2 3 4 5 6
101
4253512 77
1 2 3 4 5 6
101
4235512 77
1 2 3 4 5 6
101
4235125 77
1 2 3 4 5 6
101
N -
1
Cont’d..
Sanjay Patel
674523 14 6 3398 42
1 2 3 4 5 6 7 8
to_do
index
7
N 8 did_swap true
Cont’d..
Sanjay Patel
674523 14 6 3398 42
1 2 3 4 5 6 7 8
to_do
index
7
1
N 8 did_swap false
Cont’d..
Sanjay Patel
674523 14 6 3398 42
1 2 3 4 5 6 7 8
to_do
index
7
1
N 8
Swap
did_swap false
Cont’d..
Sanjay Patel
674598 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
1
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
674598 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
2
N 8 did_swap true
Cont’d..
Sanjay Patel
674598 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
2
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
679845 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
2
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
679845 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
3
N 8 did_swap true
Cont’d..
Sanjay Patel
679845 14 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
3
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
671445 98 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
3
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
671445 98 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
4
N 8 did_swap true
Cont’d..
Sanjay Patel
671445 98 6 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
4
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
671445 6 98 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
4
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
671445 6 98 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
5
N 8 did_swap true
Cont’d..
Sanjay Patel
671445 6 98 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
5
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
981445 6 67 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
5
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
981445 6 67 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
6
N 8 did_swap true
Cont’d..
Sanjay Patel
981445 6 67 3323 42
1 2 3 4 5 6 7 8
to_do
index
7
6
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
331445 6 67 9823 42
1 2 3 4 5 6 7 8
to_do
index
7
6
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
331445 6 67 9823 42
1 2 3 4 5 6 7 8
to_do
index
7
7
N 8 did_swap true
Cont’d..
Sanjay Patel
331445 6 67 9823 42
1 2 3 4 5 6 7 8
to_do
index
7
7
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
331445 6 67 4223 98
1 2 3 4 5 6 7 8
to_do
index
7
7
N 8
Swap
did_swap true
Cont’d..
Sanjay Patel
331445 6 67 4223 98
1 2 3 4 5 6 7 8
to_do
index
7
8
N 8
Finished first “Bubble Up”
did_swap true
For i 0 to n -2 do
{
for J 0 to n-2-i do
{
if (A[J] > A[J+1] then
{
temp A[J]
A[J] A[J+1]
A[J+1] temp
}}}
Algorithm Of Bubble sort
Sanjay Patel
Analysis
Sanjay Patel
C(n)= outer for loop * inner for loop * basic operation
n-2 n-2-i
C(n)= ∑ ∑ 1i=0 J=0
So , after deriving above formula,
C(n)= n(n-1)2
(n2 )
The time complexity of bubble sort is (n2)
Selection sort
Sanjay Patel
20 8 5 10 7
5 8 20 10 7
5 7 20 10 8
5 7 8 10 20
5 7 8 10 20
1. select the smallest element
among data[i]~ data[data.length-1];
2. swap it with data[i];
3. if not finishing, repeat 1&2
Cont’d..
Sanjay Patel
Place ith item in proper position:
temp = data[i]
shift those elements data[j] which greater than temp to right by one position
place temp in its proper position
20 8 5 10 7
20 20 5 10 7
8
temp
8
i = 1, first iteration
8 20 5 10 7---
Cont’d..
Sanjay Patel
8 20 5 10 7
8 20 20 10 7
temp
5
5
i = 2, second iteration
8 8 20 10 75
5 8 20 10 7---
Cont’d..
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5 8 20 10 7
5 8 20 20 7
temp
10
10
i = 3, third iteration
5 8 10 20 7---
Cont’d..
Sanjay Patel
5 8 10 20 7
5 8 10 20 20
5 8 10 10 20
5 8 8 10 20
7
temp
7
7
7
i = 4, forth iteration
5 7 8 10 20---
Selection sort Algorithm
Sanjay Patel
Input: An array A[1..n] of n elements.
Output: A[1..n] sorted in nondecreasing order.
{ for i 0 to n – 2 do
{ min i
for J i + 1 to n-1 {Find the i th smallest element.}
{ if A[J] < A[min] then
min J
}
temp A[i]
A[i] A[min]
A[min} temp
}
C(n)= outer for loop * inner for loop * basic operation
n-2 n-1
C(n)= ∑ ∑ 1i=0 J=i+1
The time complexity of selection sort is (n2)
Analysis of selection sort
Sanjay Patel
Radix sort
Sanjay Patel
Extra information: every integer can be represented by at
most k digits
d1d2…dk where di are digits in base r
d1: most significant digit
dk: least significant digit
Algorithm
sort by the least significant digit first (counting sort)
=> Numbers with the same digit go to same bin
reorder all the numbers: the numbers in bin 0 precede
the numbers in bin 1, which precede the numbers in bin
2, and so on
sort by the next least significant digit
continue this process until the numbers have been
sorted on all k digits
Example –Radix sort
Sanjay Patel
Least-significant-digit-first
Example: 275, 087, 426, 061, 509, 170, 677, 503
Cont’d..
Sanjay Patel
Cont’d..
Sanjay Patel
Does it work?
Clearly, if the most significant digit of a and b are different and
a < b, then finally a comes before b
If the most significant digit of a and b are the same, and the
second most significant digit of b is less than that of a, then b
comes before a.
We’ll count the total number of enqueue and
dequeue operations.
Each time through the outer for loop:
In the while loop: n elements are dequeued from q and
enqueued somewhere in array: 2*n operations
In the inner for loop: a total of n elements are
dequeued from queues in array and enqueued in q:
2*n operations
Analysis
Sanjay Patel
So, we perform 4*n enqueue and dequeue
operations each time through the outer loop
Outer for loop is executed k times, so we have
4*k*n enqueue and dequeue operations altogether
But k is a constant, so the time complexity for radix
sort is O(n)
Cont’d…..
Sanjay Patel
Insertion sort keeps making the left side of the
array sorted until the whole array is sorted.
Real life example:
An example of an insertion sort occurs in everyday life
while playing cards. To sort the cards in your hand you
extract a card, shift the remaining cards, and then
insert the extracted card in the correct place.
Insertion sort
Sanjay Patel
Sanjay Patel
Analysis
Sanjay Patel
Best case: O(n). It occurs when the data is in sorted
order. After making one pass through the data and
making no insertions, insertion sort exits.
Average case: θ(n2) since there is a wide
variation with the running time.
Worst case: O(n2) if the numbers were sorted in
reverse order.
Shell sort uses a sequence h1, h2, …, ht called the
increment sequence. Any increment sequence is fine
as long as h1 = 1 and some other choices are
better than others.
Shell sort makes multiple passes through a list and
sorts a number of equally sized sets using the
insertion sort.
Shell sort improves on the efficiency of insertion sort
by quickly shifting values to their destination.
Shell sort is also known as diminishing increment sort.
Shell sort
Sanjay Patel
Sanjay Patel
Disadvantage of Shellsort is that it is a
complex algorithm and its not nearly as
efficient as the merge, heap, and quick sorts.
The shell sort is still significantly slower than the
merge, heap, and quick sorts, but its relatively
simple algorithm makes it a good choice for
sorting lists of less than 5000 items unless
speed important. It's also an excellent choice
for repetitive sorting of smaller lists.
Pros and cons of Shell sort
Sanjay Patel
Best Case O(n) : The best case in the shell sort is
when the array is already sorted in the right order.
The number of comparisons is less.
Worst case:-The running time of Shellsort depends
on the choice of increment sequence.
The problem with Shell’s increments is that pairs of
increments are not necessarily relatively prime and
smaller increments can have little effect.
Analysis
Sanjay Patel
A heap is also known as a priority queue and can be
represented by a binary tree with the following
properties:
Structure property: A heap is a completely filled
binary tree with the exception of the bottom row,
which is filled from left to right
Heap Order property: For every node x in the
heap, the parent of x greater than or equal to the
value of x.
Heap sort
Sanjay Patel
Simple Binary run
Sanjay Patel
21 15 25 3 5 12 7 19 45 2 9
21
15 25
3 5 12 7
19 45 2 9
Array representation
Binary tree
representation
For example we have the following list of element
for sorting with heap sort method.
65,77,5,25,32,45,99,83,69,81
Now, we will build the heap structure for the given list
of element.
Heap is complete binary tree in which every parent
node has less value than child nodes.
Construction of heap structure(dec-2010)
Sanjay Patel
Sanjay PatelSanjay Patel
Constructing a heap
65 65
77
65
77 5
5
25 65
77
5
25 65
77
5
77 65
251 2 3 4
32
5
25 65
77 32 45
Now, we will swap the root node with last indexed
node. This last node of the tree will be deleted and
be placed in a queue.
Sorting with heap
Sanjay Patel
5
25 45
69 6532 99
83 77 81
Algorithm
1. Delete the root.
2. Compare the two children of the root
3. Make the lesser of the two the root.
4. An empty spot is created.
5. Bring the lesser of the two children of the empty
spot to the empty spot.
6. A new empty spot is created.
7. Continue
Delete min Algorithm
Sanjay Patel
Delete min
Sanjay Patel
1. Copy the last number to the root (i.e. overwrite the minimum element
stored there)
2. Restore the min-heap property by percolate down (or bubble down)
Cont’d
Sanjay Patel
Cont’d
Sanjay Patel
Sorting algorithm’s Complexities
Sanjay Patel
Sorting Algorithm Asymptotic Complexities
Bubble sort Best Case: O(n).
Worst Case: O (n2)
Selection sort Best Case: O(n).
Worst Case: O (n2)
Radix sort Worst Case: O(kn) where k-constant
Insertion Sort Best Case: O(n).
Worst Case: O (n2)
Shell sort Best Case: O (n)
Worst Case: O (n (log n)2
Heap Sort Best Case: O(n log n).
Worst Case: O (n log n)
All items in the left subtree are less than the root.
All items in the right subtree are greater or equal to
the root.
Each subtree is itself a binary search tree
In a binary search tree,
the left subtree contains key values less than the
root
the right subtree contains key values greater than or
equal to the root.
Binary tree search
Sanjay Patel
Cont’d..
Sanjay Patel
(a), (b) - complete and balanced trees;
(d) – nearly complete and balanced tree;
(c), (e) – neither complete nor balanced trees
We discuss four basic BST operations: search, insert,
and delete; and develop algorithms for searches,
insertion, and deletion.
• Searches
• Insertion
• Deletion
Binary search tree
Sanjay Patel
To insert data all we need to do is follow the
branches to an empty subtree and then insert the
new node.
In other words, all inserts take place at a leaf or at
a leaflike node – a node that has only one null
subtree.
Insertion
Sanjay Patel
Sanjay Patel
Cont’d..
Sanjay Patel
There are the following possible cases when we delete a node
The node to be deleted has no children. In this case, all we need to do is delete the node.
The node to be deleted has only a right subtree. We delete the node and attach the right subtree to the deleted node’s parent.
The node to be deleted has only a left subtree. We delete the node and attach the left subtree to the deleted node’s parent.
The node to be deleted has two subtrees. It is possible to delete a node from the middle of a tree, but the result tends to create very unbalanced trees.
Deletion
Sanjay Patel
Sanjay Patel
Sanjay Patel
Sanjay Patel
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