We are given a directed graph with n nodes and n-1 edges and no self loops which satisfy the below condition? We are given a directed graph with n nodes and  n-1 edges and no self loops which satisfy the below condition.
i) One of the node is reachable from every node and there is a unique path from any node to the said node.
Can we say that no node has a out-degree > 1 and the said node that is reachable from every node has outdegree=0 ,Moreover will a directed tree like structure be formed?
 A: It is given that there is a distinguished node, say $0$, in the graph such that there is a unique path from every node to $0$ (which I take to mean, a directed path).
Then we can prove that every node other than $0$ has outdegree $1$. Let $x \ne 0$ be any node. Since there is a path from $x$ to $0$, the outdegree of $x$ is at least $1$. Suppose $x$ has outdegree strictly greater than $1$, so that $y$ and $z$ are two nodes to which $x$ is adjacent (i.e., there are arcs from $x$ to both $y$ and $z$). Now, there are paths, say $P_y$ and $P_z$ from $y$ and $z$ respectively to node $0$. Then we obtain two distinct paths $xP_y$ and $x P_z$ are from $x$ to $0$, which contradicts the given condition.
If by "path", you mean a walk (i.e., if vertices are allowed to repeat), then you can conclude that node $0$ has outdegree $0$. For, if there is an arc from $0$ to any node $x$, we obtain a directed cycle by adjoining the arc $0 \to x$ with the path from $x$ to $0$. The existence of such a cycle contradicts uniqueness of walks from each node to $0$ (since we can traverse a path from any node $y$ to the node $0$ and then traverse the cycle from $0$ to itself repeatedly, to obtain any number of distinct walks from $y$ to $0$).
Otherwise, we can also use the condition on the number of nodes and edges to conclude the same. Again, if node $0$ has non-zero outdegree, then there is a directed cycle in the graph. Now, consider the underlying undirected graph, in which each arc is replaced by an undirected edge. Clearly, it is connected (since there is path from every node to node $0$). If the number of edges is $n - 1$ (where $n$ is the number of nodes), the graph is a tree and is therefore acyclic, which is a contradiction. Thus, node $0$ must have outdegree $0$ in the original directed graph.
It is obvious from the above that the graph is indeed a directed tree — in fact, it is an in-tree.
