PT07Y - Is it a tree

You are given an unweighted, undirected graph. Write a program to check if it's a tree topology.


The first line of the input file contains two integers N and M --- number of nodes and number of edges in the graph (0 < N <= 10000, 0 <= M <= 20000). Next M lines contain M edges of that graph --- Each line contains a pair (u, v) means there is an edge between node u and node v (1 <= u,v <= N).


Print YES if the given graph is a tree, otherwise print NO.


3 2
1 2
2 3


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RADHE SHYAM LODHI: 2016-01-12 05:10:29

AC union set 0.00 ..

dokz: 2016-01-06 06:33:31

Got AC. Just run the single DFS:
1) Check for loops - grey (visited, but not left) vertex found - loop, output "NO"
2) Check for single connected component - after DFS all vertices should be colored black (visited and left)

Deepak : 2015-12-30 12:34:06

thanks @rahul for your hint

AASHISH KUMAR: 2015-12-25 10:01:02

AC in one go :) yeyii

rahul: 2015-12-23 19:29:15

i think this question can be solved easily if you focus on checking how many connected component it has if it has one connected component and edges=nodes-1 it will b accepted

ankurverma1994: 2015-12-23 15:22:25

Got AC with Union Find in 1st Attempt in Java. Don't know why its showing WA when using DFS in Java and getting AC(using same DFS code as in Java) in C++ with same logic... :(

MAYANK NARULA: 2015-12-08 09:41:20

Well This problem gave me some WAs .!!!!.... Maybe Graph has Self - loops at some vertices..

sarthak_8: 2015-12-04 22:14:22

Yeaahh !! Finally got an AC After 3 WA. People who are new to graph (as I was) should try to look how to implement a graph and then check the ewquired conditions for a graph to be a tree. There are 3 conditions, If you want to learn try to check all 3 and not the only easiest one. Geeksforgeeks has a similar question, you can view it to understand the working of graphs.

karthik1997: 2015-11-21 14:23:58

For people
Union set gets you done in 0.0 s
checking conditions are for tree are
The inputs are integer's from 1 to n
1. input must be in order => in the m line 2 3 is allowed but 3 2 in input is not allowed (i got this doubt so i wasted an hour thinking about this or else it was a matter of 5 mins . ) NO
2. if a b are having edge . b should not be connected to any NO
3. if it is not connected , check for repetition of b in parents of a 's .if b is there ,it is cycle NO , else join them
4. at the end ,loop and find if there are any elements with no , count them . if it is not 1 then NO
5. at the end the answer is YES if all the 4 fail
Have fun very easy one ac in one go :)

Last edit: 2015-11-21 14:34:41
garmel: 2015-11-13 21:27:43

I think if you aren't confortable with graphs, you should search for a solution and understand it by using the hand-execution on a paper...

Added by:Thanh-Vy Hua
Time limit:0.166s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ERL JS
Resource:Co-author Amber