枚举树中的所有路径 [英] Enumerating all paths in a tree

问题描述

我想知道如何最好地实现树数据结构，以便能够枚举所有级别的路径.让我用下面的例子来解释它:

I was wondering how to best implement a tree data structure to be able to enumerate paths of all levels. Let me explain it with the following example:

     A
   /   
   B    C
   |    /
   D   E  F

我希望能够生成以下内容:

I want to be able to generate the following:

A
B
C
D
E
F
A-B
A-C
B-D
C-E
C-F
A-B-D
A-C-E
A-C-F

截至目前，我正在对使用字典构建的数据结构执行不同深度的深度优先搜索，并记录看到的唯一节点，但我想知道是否有更好的方法来执行这种搜索遍历.有什么建议吗?

As of now, I am executing a depth-first-search for different depths on a data structure built using a dictionary and recording unique nodes that are seen but I was wondering if there is a better way to do this kind of a traversal. Any suggestions?

推荐答案

每当你在树上发现问题时，就使用递归 :D

Whenever you find a problem on trees, just use recursion :D

def paths(tree):
  #Helper function
  #receives a tree and 
  #returns all paths that have this node as root and all other paths

  if tree is the empty tree:
    return ([], [])
  else: #tree is a node
    root = tree.value
    rooted_paths = [[root]]
    unrooted_paths = []
    for subtree in tree.children:
        (useable, unueseable) = paths(subtree)
        for path in useable:
            unrooted_paths.append(path)
            rooted_paths.append([root]+path)
        for path in unuseable:
            unrooted_paths.append(path)
    return (rooted_paths, unrooted_paths)

def the_function_you_use_in_the_end(tree):
   a,b = paths(tree)
   return a+b

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