问题
I'm looking for some algorithm which will help me find all possible paths in a graph. Everything I found so far is not fully satisfying.
Let's imagine we have a graph (tree) like this one:
And let's use some algorithm like Breadth-First Search or Depth-First Search. In return we'll get something like
1, 2, 4, (2), 5, (2), 6, (2), (1), 3, 7, 8, (7), 9
Which is how we go through this tree and this is not what I'm looking for. I'd love to get all paths, like:
1
1, 2
1, 2, 4
1, 2, 5
1, 2, 6
1, 3
1, 3, 7
1, 3, 7, 8
1, 3, 7, 9
The thing is that I want just to specify root
node and algorithm should be able to provide me with all possible paths of any length.
So far, the simple code I have looks like:
func dfs(_ graph: Graph, source: Node) -> [String] {
var nodesExplored = [source.label]
source.visited = true
for edge in source.neighbors {
if !edge.neighbor.visited {
nodesExplored += dfs(graph, source: edge.neighbor)
}
}
return nodesExplored
}
回答1:
Just fold the result and count the new possibilities: Result = 9 (you forgott the path [1] )
n0 n1 n2 n3
1, 2, 4, +3
(2), 5, +1
(2), 6, +1
(2), +0
(1), 3, 7, 8, +3
(7), 9 +1
回答2:
You can use this algorithm on a binary tree to print out all of its root-to-leaf paths. The function treePaths
traverses the nodes depth-first (DFS), pre-order, recursively.
treePaths(root, path[1000], 0) // initial call, 1000 is a path length limit
// treePaths traverses nodes of tree DFS, pre-order, recursively
treePaths(node, path[], pathLen)
1) If node is not NULL then
a) push data to path array:
path[pathLen] = node->data.
b) increment pathLen
pathLen++
2) If node is a leaf node, then print the path array, from 0 to pathLen-1
3) Else
a) Call treePaths for left subtree
treePaths(node->left, path, pathLen)
b) Call treePaths for right subtree.
treePaths(node->right, path, pathLen)
来源:https://stackoverflow.com/questions/41248914/finding-all-possible-paths-in-graph