Is a directed graph uniquely determined by the in/out degree of each node? I never really thought of this problem. If we have two directed graphs $A$ and $B$ with the same set of nodes $V$, and we know that the in/out degree of each node is the same in the $A$ and $B$, is it possible for $A$ and $B$ to have a different set of edges? In other words, does the in/out degree of each node guarantee the uniqueness of a graph?
I have tried to find a counter-example of two graphs which demonstrate this property (each node has same in/out degree in $A$ and $B$, but the edge set of the two graphs is different), but have come up empty handed thus far. Perhaps next would be finding a proof, or a counter-example.
As a further edit, the graphs I have in mind are connected.
 A: These two graphs are a counterexample to your conjecture; in each graph, all four vertices have indegree 1 and outdegree 1: 
To make the example connected, just embed these two graphs into some larger graph.  For example:

Here the I have added two red vertices and some additional edges.  Each result has corresponding vertices with the same indegree and outdegree.
It should be clear that by adding different numbers of red vertices or edges to or from them, the basic example can be extended to produce many different examples.
A: Consider the graph on A, B, C with directed edges AB, BC, CA, and the graph with AC, BA, CB. 
If you want the graphs nonisomorphic, take AB, BC, DC, ED, FC versus AC, DC, BD, EB, FC. 
A: Take as $A$ two triangles, with vertices $RST$ and $UVW$. Connect $RST$ cyclically, also $UVW$. Join $R$ and $U$ by a directed edge.
Take as $B$ a hexagon $RSTUVW$, and join $R$ and $U$ by a directed edge. 
Graphs $A$ and $B$ are connected. The degree sequences of $A$ and $B$ are the same. But the graphs are not isomorphic, since the first can be disconnected by removing an edge, and the second cannot be. 
Remark: A difficult problem in graph theory is to find an efficient algorithm for determining whether two graphs are isomorphic. An obvious necessary condition is that the degree sequences match, but that condition is far from sufficient. 
A: Definitely not. 


*

*First, whenever the graph is directed or undirected does not matter much, since you can put an edge in two directions.

*Any planar graph can be made into cubic planar graph by expanding nodes into cycles. In such case there are many which are not the same or not even isomorphic.

*To give a simpler example, this is not true for trees, e.g. (with bidirectional edges)
$$((1,2,3),(4,5,6)) \text{ and } ((a,b),c,(d,e,f)).$$


I hope this helps $\ddot\smile$
