r - igraph R中从根到叶的有向 TreeMap 中的所有路径

Question

给定是一棵树:

library(igraph)

# setup graph
g= graph.formula(A -+ B,
                 A -+ C,
                 B -+ C,
                 B -+ D,
                 B -+ E
)
plot(g, layout = layout.reingold.tilford(g, root="A"))

顶点 "A"是树的根，而顶点 "C", "D", "E"被视为终端叶。

问题:

任务是找到根和叶之间的所有路径。我失败了以下代码，因为它只提供最短路径:

# find root and leaves
leaves= which(degree(g, v = V(g), mode = "out")==0, useNames = T)
root= which(degree(g, v = V(g), mode = "in")==0, useNames = T)

# find all paths
paths= lapply(root, function(x) get.all.shortest.paths(g, from = x, to = leaves, mode = "out")$res)
named_paths= lapply(unlist(paths, recursive=FALSE), function(x) V(g)[x])
named_paths

输出:

$A1
Vertex sequence:
[1] "A" "C"

$A2
Vertex sequence:
[1] "A" "B" "D"

$A3
Vertex sequence:
[1] "A" "B" "E"

问题:

如何找到包括顶点序列在内的所有路径:"A" "B" "C" ?

我的理解是，缺失的序列 "A" "B" "C"不是由 get.all.shortest.paths() 提供的作为来自 "A" 的路径至 "C"通过顶点序列:"A" "C" (在列表元素 $A1 中找到)较短。所以igraph工作正常。
尽管如此，我正在寻找一种代码解决方案，以 R list 的形式获取从根到所有叶子的所有路径.

评论:

我知道对于大树，覆盖所有组合的算法可能会变得昂贵，但我的实际应用程序相对较小。

Answer 1

根据 Gabor 的评论:

all_simple_paths(g, from = root, to = leaves)

产量:

[[1]]
+ 3/5 vertices, named:
[1] A B C

[[2]]
+ 3/5 vertices, named:
[1] A B D

[[3]]
+ 3/5 vertices, named:
[1] A B E

[[4]]
+ 2/5 vertices, named:
[1] A C