Why dfs using stack is not giving correct result - graph

Question

See DFS image Here
I am using stack to print sequence of dfs. According to input and that image of graph, sequence is 1 2 4 8 5 6 3 7 . But My code is giving output as 1 2 4 8 7 6 5 3 . Can anyone explain how can i fix it??
Input:
8 10
1 3
1 2
2 5
2 4
3 7
3 6
4 8
5 8
6 8
7 8
Correct Output:
Sequence: 1 2 4 8 5 6 3 7
My Code :
#include <bits/stdc++.h>
using namespace std;
vector<int>edges[100];
stack<int>q;
vector<int>item;
int level[100],parent[100],visited[100],tn;
void dfs(int s)
{
int j,k,fr;
q.push(s);
level[s]=0;
for(j=1;j<=tn;j++)
{
visited[j]=0;
}
visited[s]=1;
while(!q.empty())
{
fr=q.top();
q.pop();
item.push_back(fr);
for(k=0;k<edges[fr].size();k++)
{
if(visited[edges[fr][k]]==0)
{
q.push(edges[fr][k]);
//cout<<"Pushed="<<fr<<"="<<edges[fr][k];
visited[edges[fr][k]]=1;
}
}
//cout<<endl;
}
}
int main()
{
int i,e,p,n,u,v,f,m;
cin>>tn>>e;
for(i=1;i<=e;i++)
{
cin>>u>>v;
edges[u].push_back(v);
edges[v].push_back(u);
}
dfs(1);
cout<<"Sequence="<<endl;
for(m=0;m<item.size();m++)
{
cout<<item[m];
}
return 0;
}
My Code is showing this output: 1 2 4 8 7 6 5 3

Answer 1

The marking of the nodes as visited in the implementation contains a bug; the function can be rewritten as follows.
void dfs(int s)
{
int j, k, fr;
q.push(s);
level[s] = 0;
for (j = 1; j <= tn; j++)
{
visited[j] = 0;
}
while (!q.empty())
{
fr = q.top();
q.pop();
if (0 == visited[fr])
{
visited[fr] = 1;
item.push_back(fr);
for (k = 0; k < edges[fr].size(); k++)
{
q.push(edges[fr][k]);
}
}
}
}
In this version, a node gets marked only if it is taken from the stack. Note that a check whether the node has been already visited is necessary, as a node on the stack might be visited by a later iteration. This implementation yields the sequence
1 2 4 8 7 3 6 5
which, however, is not the one described as desired solution. However, note that without additional tie-breaking rules, the DFS algorithm permits some ambiguity in the sequence of visits. The sequence
1 2 4 8 5 6 3 7
can be generated by pushing a neighbor with smallest id to the stack last, causing it to be visited in the next iteration.