You are given an directed graph $G = (V , E)$ with positive edge capacities, a source vertex $s \in V$, and a target vertex $t \in V \setminus \{s\}$. $G$ is guaranteed to have the following additional property: The graph obtained by deleting $t$ from $G$ has exactly one path from $s$ to each other vertex in $V \setminus \{t\}$. How can we compute the maximum flow from $s$ to $t$ in $G$ in time $O(|V|+|E|)$?
First, I read this post, and i think start from $s$ then run BFS, when we visit each node then set it to minimum flow that we find from $s$ to that vertex. Do this until we achieve vertices that have edges to $t$ and push each flow inside that vertices to $t$ this way we have an algorithm with linear time. Is argument correct?