Counting number of left and right nodes without Using cursors.
Hi!
I'm using sql server 2000, I need to count number of nodes in left and
right side of a root node....
suppose i've entries in table as
node
parent_node Position node_name
1 0
'L' abc
2 1 'L'
b1
3 1 'R' b2
4
2 'L' c1
5
2 'R' c2
6
3 'L' d1
if i supplied node '1' then
it should Give result as
left count=3 and
right count =2
actually what i
want is this..
For my entries tree will be like this..
1(abc)
/ \
/ \
(b1) 2 3(b2)
/\ / \
/ \ / \
(c1)4 5(c2) 6(d1)
Now if i give i/p as 1 then
eft count=3 (b1,c1,c2) and
right count =2 (b2,d1)
if the tree is like this...
1(abc)
/ | \
/ | \
(b1) 2 |
3(b2)
/\ | /
/
\ | /
(c1)4 5(c2)| 6(d1)
| / \
| / \
| (d2)7 8(e1)
the middle line is just imaginary to differentiate left and right nodes for node '1(abc)'
Now if i give i/p as 1 then
eft count=3 (b1,c1,c2) and
right
count =4(b2,d1,d2,e1)
I DONT WANT TO USE CURSORS BECAUSE I'VE SOLUTION WITH CURSOR BUT THERE ARE THOUSANDS OF RECORDS IN THE TABLE ,SO EXECUTION TIME OF PROCEDURE IS V VERY HIGH SOMETIMES IT STRUCK SO PLEASE SUGGEST ME SOMETHING......
Please....Please help me out i'm strucked from many days...Thanx in advance..