Balanced trees ppt in DS

1. Active Learning Assignment
Data Structures
Prepared By:-Darji meet B (160120107025)
Topic :- Balanced trees
Gandhinagar Institute of Technology

2. What is Balanced
Tree ?

3. A binary tree is balanced if for each node it
holds that the number of inner nodes in the
left subtree and the number of inner nodes in
the right subtree differ by at most 1.

4. Example
7
8
5
126
11
4
9
13
10
T3
T1
T4
T2
X
Y
Z

5. (b)
Z
X
Y
T1 T2 T4
115
4 9
8
6 12
7
1310
T3

6. BALANCED TREES ALGORITHM
Procedure:-
CREATE_LEAF(NAME,INFO,NEW)
1[Create a new leaf node and initialize]
NEW NODE
LPTR(NEW) RPTR(NEW) NULL
BI(NEW) ‘B’
K(NEW) NAME
DATA(NEW) INFO
2[Finished]
Return

7. Function BALANCED_INSEAT(HEAD, NAME,INFO)
1.[Is this a first insertion?]
If LPTR(HEAD)=HEAD
then Call CREATE_LEAF(NAME,INFO,NEW)
LPTR(HEAD) NEW
Return(NEW)
2.[Initialize]
LEVEL 0
DIRECTION(LEVEL) ‘L’
PATH[LEVEL] HEAD
T LPTR(HAND)
3.[Compare until found or inserted]
Repeat step 4 forever
4.[Compare and insert, if required]
If NAME<K(T)

8. then If LPTR(T)≠ NULL
then LEVEL LEVEL+1
PATH[LEVEL] T
DIRECTION[LEVEL] ‘L’
T LPTR(T)
else Call CREATE_LEAF(NAME,INFO,NEW)
LPTR(T) NEW
LEVEL LEVEL +1
PATH[LEVEL] T
DIRECTION[LEVEL] ‘L’
Exitloop
else If NAME>K(T)
then If RPTR(T)≠NULL
then LEVEL LEVEL+1
PATH[LEVEL] T
DIRECTION[LEVEL] ‘R’
T RPTR(T)
Else call CREATE_LEAF(NAME,INFO,NEW)
RPTR(T) NEW
LEVEL LEVEL+1

9. PATH[LEVEL] T
DIRECTION[LEVEL] ‘R’
Exitloop
else (A match; NAME=K(T))
write(‘ITEM ALREADY THERE’)
Return(NULL)
5.[search for an unbalanced node]
MARK 0
Repeat for l= LEVEL,LEVEL-1,….,1
P PATH[1]
if Bl(P) ≠ B
then MARK 1
Exitloop
6.[Adjust balance indicators]
Report for l= MARK+1,MARK+2,….LEVEL
lf NAME<K(PATH[L])
then Bl(PATH[l]) ‘L’
else Bl(PATH[l]) ‘R’

10. 7.[Is there a critical node?]
If MARK=0
then Return(NEW)
D DIRECTION[MARK]
X PATH[MARk]
Y PATH[MARK+1]
If Bl(X)≠D
then (the node was heavy and now becomes balanced)
Bl(X) ‘B’
Return(NEW)
8.[Rebalancing: case1]
If Bl(Y)=D
then (the node was heavy and now becomes balanced)
If D=‘L’
then LPTR(X) RPTP(Y)
RPTR(Y) X

11. else RPTR(X) LPTR(Y)
LPTR(Y) X
Bl(X) Bl(Y) ‘B’
F PATH[MARK-1]
If X=LPTR(F)
then LPTR(F) Y
else RPTR(F) Y
Return(NEW)
9.[Rebalancing tree: case 2]
(a) (Change structure links)
If D=‘L’
then Z RPTR(Y)
RPTR(Y) LPTR(Z)
LPTR(Z) Y
LPTR(X) RPTR(Z)
RPTR(Z) X

12. else Z LPTR(Y)
LPTR(Y) RPTR(Z)
RPTR(Z) Y
RPTR(X) LPTR(Z)
LPTR(Z) X
F PATH[MARK-1]
If X = LPTR(F)
then LPTR(F) Z
else RPTR(F) X
(b) (Chang balance indicators)
If Bl(Z)=D
then Bl(Y) Bl(Z) ‘ B’
If D =‘L’
then Bl(X) ‘R’
else Bl(X) ‘L’

13. else If Bl(Z)=B
then Bl(X) Bl(Y) Bl(Z) B
else Bl(X) Bl(Z) B
Bl(Y) D
Return(NEW)