Python生成节点间所有路径程序报错求助：UnboundLocalError

问题原因 #

你的代码中，当遍历到已经存在于当前path中的节点时，newpaths变量未被初始化就直接在for循环中被引用。例如递归到节点e时，它的邻接节点是b，而b已经在当前路径里，此时if node not in path条件不成立，newpaths没有被赋值，后续执行for newpath in newpaths就会触发未定义错误。

修复后的代码 #

写法一：调整代码块范围

graph = { 
    "a" : ["b","c"],
    "b" : ["a", "d"],
    "c" : ["a", "d"],
    "d" : ["e"],
    "e" : ["b"]
}

def find_all_paths(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return [path]
    paths = []
    for node in graph[start]:
        if node not in path:
            newpaths = find_all_paths(graph, node, end, path)
            for newpath in newpaths:
                paths.append(newpath)
    return paths

print(find_all_paths(graph, "b","e"))

写法二：用extend简化逻辑（更推荐）

graph = { 
    "a" : ["b","c"],
    "b" : ["a", "d"],
    "c" : ["a", "d"],
    "d" : ["e"],
    "e" : ["b"]
}

def find_all_paths(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return [path]
    paths = []
    for node in graph[start]:
        if node not in path:
            # 直接将递归返回的路径列表合并到paths中
            paths.extend(find_all_paths(graph, node, end, path))
    return paths

print(find_all_paths(graph, "b","e"))

修复说明 #

• 写法一：将for newpath in newpaths移到if条件块内部，确保只有节点不在路径中时才执行递归和路径添加，避免引用未定义变量。
• 写法二：使用extend方法直接合并递归返回的路径列表，彻底消除单独定义newpaths的需求，代码更简洁且从根源避免错误。

运行修复后的代码，会输出正确路径：[['b', 'a', 'c', 'd', 'e'], ['b', 'd', 'e']]

内容的提问来源于stack exchange，提问作者SumSum