Some how I found that we need to run bfs twice to get the largest distance
between any two nodes of a tree. Please explain me how it works.
regards,
avinash
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Is this similar to finding the diameter of a tree(longest disteance between
two leaves) ?? If yes than visit this link
http://www.cs.duke.edu/courses/spring00/cps100/assign/trees/diameter.html
On Sat, Jun 23, 2012 at 2:44 PM, Avinash jain.av...@gmail.com wrote:
Some how I found that we need to
Is longest path between two node in a tree is equal two logest path between
two leaves??
On Sat, Jun 23, 2012 at 5:35 PM, Navin Kumar algorithm.i...@gmail.comwrote:
Is this similar to finding the diameter of a tree(longest disteance
between two leaves) ?? If yes than visit this link
no this problem is different from finding the longest path which is
actually maximum number of nodes covered in the longest path
On Sat, Jun 23, 2012 at 5:45 PM, Navin Kumar algorithm.i...@gmail.comwrote:
Is longest path between two node in a tree is equal two logest path
between two leaves??
What I can think
is case is :
10
/ \
6 13
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
so from a-b is
a-7-4-2-5-8-b
1- Left Tree then
2- Right Tree
add them
On Sat, Jun 23, 2012 at 3:49 PM, Kumar Vishal kumar...@gmail.com wrote:
What I can think
What I can think
is case is :
1
/ \
2 3
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
so from a-b is
a-7-4-2-5-8-b
On Sat, Jun 23, 2012 at 2:44 PM, Avinash jain.av...@gmail.com wrote:
Some how I found that we need to run bfs twice to get the largest distance
you are just finding the diameter of the tree. Remember the graph I am
talking about is weighted. So if two adjacent vertices has infinite weight
than that would be the longest distance between any two vertices in the
graph.
e.g. In you diagram suppose the sum of weight of the edges from a-b is
