使用python的DFS incorerct输出

Question

我问了一个类似的问题，但在合并更改后也提供了不正确的输出，请帮忙！

上一个问题 Depth First Search in python Error : Key Error 7

当前代码：

output=[]
graph = {
           1:[2,3],
           2:[4,5],
           3:[6,7],
           4:[],
           5:[],
           6:[],
           7:[]
        }

def dfs(graph,root):
    stack=[]
    visited=set()

    stack.append(root)
    output.append(str(root))
    visited.add(root)

    while not(stack==[]):
        for item in graph[root]:

            if item not in visited:
                stack.append(item)
                visited.add(item)
                output.append(str(item))

            if set(graph[item]).union(visited)==visited:
                stack.pop(-1)
                if not(stack==[]):
                    root=stack[len(stack)-1]
                else:
                    break
                continue

            root=item

dfs(graph,1)
print(" ".join(output))

Answer 1

你的工作太辛苦了...... DFS易于实施

output=[]
graph = {
           1:[2,3],
           2:[4,5],
           3:[6,7],
           4:[],
           5:[],
           6:[],
           7:[]
        }

def dfs(graph,root):
    stack=[]
    visited=set()
    stack.append(root)


    while stack:
        node = stack.pop() #remove last
        if node in visited:
           continue
        visited.add(node)
        output.append(str(node))
        children = graph[node]
        stack.extend(children)



dfs(graph,1)
print(" ".join(output))

如果您希望输出完全匹配，则需要更改

stack.extend(children) 

到

stack.extend(children[::-1])