In directed graph example

N=3 ,M=1. N-Number of vertices , M - Edges

Graph is only one edge

1-3

Does it have euler path/circuit ?

As it is not connected so no eulerian path/cycle exits ? But answer is 1-3

Can anyone explain this?

In directed graph example

N=3 ,M=1. N-Number of vertices , M - Edges

Graph is only one edge

1-3

Does it have euler path/circuit ?

As it is not connected so no eulerian path/cycle exits ? But answer is 1-3

Can anyone explain this?

The necessary condition for a graph to contain Euler path is that it must have exactly two odd vertices i.e. vertices of odd degree.

So, as in the example, only vertices 1 and 3 have odd degree i.e. 1 , therefore the graph contains an Euler path viz. 1-3

Here the graph is directed. The Condition of odd degree is for undirected graph right?

For Euler path, it has to be connected. Here in example vertex 2 is disconnected from graph. So No Euler path right?

Or should we have to check Connectivity only for non-zero degree vertices?