= Solution
\b[No: equal-length shortest paths permit different trees.] Consider edges $s\to a,s\to b,a\to t,b\to t$, all of weight one. Both routes to $t$ have length two. A <shortest-path tree> may therefore choose either $a$ or $b$ as the predecessor of $t$.
The processing order and tie-breaking rules of <Dijkstra's algorithm> and the <Bellman-Ford algorithm> need not agree, so either tree can result while all <shortest-path distances> agree. If every reachable vertex has a unique shortest path from $s$, then both correct algorithms must select those same paths.
Back to article page
