Question

我的结构如下:

config
  |-- groups
    |-- rootgroup
    |-- group1 (includes rootgroup)
    |-- group2 (includes group1)
    |-- group3 (includes rootgroup)
  |-- users
    |-- Fred (includes group3 and group2)

因此,Fred的继承树将像:

    _Fred_
   v      v
group2  group3
   v      v
group1    v
   v     /
rootgroup

我需要一种算法,从树底部开始(例如,它将是根基组——组1 - 第二组——组3;组1超越了根基组,组2大于锡石组1等)并找到复读联系(例如,如果根群包含第2组的话),此外,它必须找到再入侵的循环(......和;组2 - 和组;组1-和组;根组-组;组2 ->组......)。

有利可图的语言是假的,但任何意愿都是如此。

感谢。

Answer 1

在阅读直观图像之后,我提出了以下解决办法:

def getNodeDepsTree(self, node, back_path=None):
    """Return whole dependency tree for given node"""
    # List of current node dependencies
    node_deps = []

    if not back_path:
        back_path = []

    # Push current node into return path list
    back_path.append(node)

    # Get current node dependencies list
    deps = getNodeDeps(node)

    for dep in deps:
        # If dependency persist in list of visited nodes - it s a recursion
        if dep in back_path:
            pos = back_path.index(dep)
            self.log.error("Recursive link detected.  " + node
                    + " , contains a recursive link to  " + dep
                    + " . Removing dependency. Recursive loop is:
	"
                    +  
	 .join(back_path[pos:]))
            continue
        # Recursively call self for child nodes
        node_deps.extend(self.getNodeDepsTree(dep, back_path))

    # Finally add current node as dependency
    node_deps.append(node)

    # Remove current node from list of visited nodes and return
    back_path.pop()

    return node_deps

友情链接