Little Help in Euler Path in Directed Graph Concept?

1

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?

2

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

3

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

4

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

5

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

6

My bad I didn’t see it’s directed.

I’ve found a Wiki article on its properties w.r.to directed graph since I wasn’t clear with it myself:

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