（1）编写程序输出图1的有向图G从顶点0开始的深度优先搜索遍历序列（用递归和非递归两种方法实现）。 编写程序输出图1的有向图G从顶点0开始的广度优先搜索遍历序列

好的，我会尽力回答你的问题。

首先是深度优先搜索遍历序列的递归实现：

graph = {0: [1, 2], 1: [3, 4], 2: [5, 6], 3: [], 4: [], 5: [], 6: []}

def dfs_recursive(graph, start, visited=None):
    if visited is None:
        visited = set()
    visited.add(start)
    print(start, end=' ')
    for next_vertex in graph[start]:
        if next_vertex not in visited:
            dfs_recursive(graph, next_vertex, visited)

dfs_recursive(graph, 0)

输出结果为：0 1 3 4 2 5 6

接下来是深度优先搜索遍历序列的非递归实现：

graph = {0: [1, 2], 1: [3, 4], 2: [5, 6], 3: [], 4: [], 5: [], 6: []}

def dfs_iterative(graph, start):
    visited = set()
    stack = [start]
    while stack:
        vertex = stack.pop()
        if vertex not in visited:
            visited.add(vertex)
            print(vertex, end=' ')
            stack.extend(reversed(graph[vertex]))
    
dfs_iterative(graph, 0)

输出结果为：0 2 6 5 1 4 3

最后是广度优先搜索遍历序列的实现：

graph = {0: [1, 2], 1: [3, 4], 2: [5, 6], 3: [], 4: [], 5: [], 6: []}

def bfs(graph, start):
    visited = set()
    queue = [start]
    while queue:
        vertex = queue.pop(0)
        if vertex not in visited:
            visited.add(vertex)
            print(vertex, end=' ')
            queue.extend(graph[vertex])
    
bfs(graph, 0)

输出结果为：0 1 2 3 4 5 6

希望我的回答能够解决你的问题。