python-查找树中两个节点之间的(保证唯一)路径

我有一个(可能)简单的图形遍历问题.我是使用networkx作为图形数据结构的图形新手.我的图总是这样：

             0
      1              8
   2     3       9      10
 4  5   6 7    11 12  13  14

我需要返回从根节点到给定节点的路径(例如,path(0,11)应该返回[0,8,9,11]).

我有一个解决方案,它通过传递一个列表来工作,该列表会随着树的移动而增大和缩小,以跟踪路径的样子,最终在找到目标节点时返回：

def VisitNode(self, node, target, path):
    path.append(node)
    # Base case. If we found the target, then notify the stack that we're done.
    if node == target:
        return True
    else:
        # If we're at a leaf and it isn't the target, then pop the leaf off
        # our path (backtrack) and notify the stack that we're still looking
        if len(self.neighbors(node)) == 0:
            path.pop()
            return False
        else:
            # Sniff down the next available neighboring node
            for i in self.neighbors_iter(node):
                # If this next node is the target, then return the path 
                # we've constructed so far
                if self.VisitNode(i, target, path):
                    return path
            # If we've gotten this far without finding the target, 
            # then this whole branch is a dud. Backtrack
            path.pop()

我感到骨子里没有必要传递这个“路径”列表…我应该能够使用调用堆栈来跟踪该信息,但是我不知道如何…有人可以启发一下我如何使用堆栈来递归地解决路径问题？

解决方法:

通过在失败时返回None,在成功时返回部分路径,可以避免绕过路径.这样,您无需保留从根到当前节点的某种“面包屑路径”,而是仅在找到目标后才构造从目标到根的路径.未经测试的代码：

def VisitNode(self, node, target):
    # Base case. If we found the target, return target in a list
    if node == target:
        return [node]

    # If we're at a leaf and it isn't the target, return None 
    if len(self.neighbors(node)) == 0:
        return None

    # recursively iterate over children
    for i in self.neighbors_iter(node):
        tail = self.VisitNode(i, target)
        if tail: # is not None
            return [node] + tail # prepend node to path back from target
    return None #none of the children contains target

我不知道您正在使用的图形库,但是我假设即使叶子也包含neighbours_iter方法,该方法显然不应该为叶子产生任何子级.在这种情况下,您可以省略对叶子的显式检查,从而使其变得更短：

def VisitNode(self, node, target):
    # Base case. If we found the target, return target in a list
    if node == target:
        return [node]
    # recursively iterate over children
    for i in self.neighbors_iter(node):
        tail = self.VisitNode(i, target)
        if tail: # is not None
            return [node] + tail # prepend node to path back from target
    return None # leaf node or none of the child contains target

我还删除了其他else语句,因为如果要从函数返回,则在true的内部.这是常见的refactering pattern(有些老派人士不喜欢).这消除了一些不必要的缩进.

标签：graph-traversal,networkx,python
来源： https://codeday.me/bug/20191030/1967329.html