I am Learning Articulation points in Graph thoery which is an application of DFS.I wrote the code by taking reference from GeeksforGeeks [Atriculation Points].(http://www.geeksforgeeks.org/articulation-points-or-cut-vertices-in-a-graph/)

void dfs(int u)
{
	static int time=0;
	int child=0;
	vis[u]=1;
	dis_t[u]=low[u]=++time;
	for(int i=0;i<G[u].size();i++)
	{
		int v=G[u][i];
		if(!vis[v])
		{
			child++;
			parent[v]=u;
			dfs(v);
			low[u]=min(low[u],low[v]);
			if(parent[u]==-1 and child>1)
			{	
				AP[u]=1;
			}
			if(parent[u]!=-1 and low[v]>=dis_t[u])
				AP[u]=1;
		}
		else if(v!=parent[u])
		 	low[u]=min(low[u],dis_t[v]);
	}
}

I came across two conditions.
a vertex u is articulation point if one of the following two conditions is true.
1) u is root of DFS tree and it has at least two children.
2) u is not root of DFS tree and it has a child v such that no vertex in subtree rooted with v has a back edge to one of the ancestors (in DFS tree) of u.

First condition is clear,but I can,t understand the second one.Hoping some help:)