Assume you have a simple, infinite graph $G$ with bounded degree (there is an upper bound for the degree of the nodes). Choose an arbitrary vertex $x\in V(G)$ and consider $$ G_{n}:=\{x\in G:d(x_0,x)\leq n\} $$ with the graph metric (hop metric). Assume that each pair of nodes is communicating a unit load of information and the load goes through the minimum path between nodes (if there is more than one minimum path we choose one arbitrarily). The total traffic in $G_{n}$ is equal to $\frac{N(N-1)}{2}$ where $N=N(n)=|G_{n}|$.

Given a node $v\in G_{n}$ we define $T_{n}(v)$ as the total traffic generated in $G_{n}$ passing through $v$. In other words, $T_{n}(v)$ is the sum off all the geodesic paths in $G_{n}$ which are carrying traffic and contain the node $v$.

If the graph $G$ is planar and it has exponential growth $|G_{n}|=K\exp(\lambda n)$ for $n$ sufficiently large, then it is not difficult to prove that there are nodes in $G_{n}$ such that $$ T_{n}(v)\geq C\frac{N^2}{\log(N)} $$ for $n$ sufficiently large.

**My question is the following**

Is the same true if we remove the planar condition but we keep the exponential growth? My intuition is that the answer is no but I can't find a counterexample.