When extracting path from start to destination node ,iterate through vector in descending order using the returned vector then it is possible to find path in O(n)-Time while otherwise in O(n**2)!

So:
let mut node: &Node = dest;

for i in vec.len() to 0 /* from top to bottom */ {
if searchedVector = node {
node = searchedVector.prev;
// One node of path found!
}
}

Same procedure is also possible in ascending order using start node. This exploits the fact that the dijkstra algorithm proceeds level by level, i.e. the nodes of a path always appear in the vector in the order in which they were viewed.