تمثيل الرسوم والبحث بالعرض والعمق

from collections import deque

def bfs_distances(graph, source):
    if source not in graph:
        raise ValueError("Unknown source")
    distances = {source: 0}
    queue = deque([source])
    while queue:
        vertex = queue.popleft()
        for neighbor in graph[vertex]:
            if neighbor not in distances:
                distances[neighbor] = distances[vertex] + 1
                queue.append(neighbor)
    return distances

graph = {"A": ["B", "C"], "B": ["D"], "C": ["D"], "D": [], "E": []}
print(sorted(bfs_distances(graph, "A").items()))

[('A', 0), ('B', 1), ('C', 1), ('D', 2)]

تُدرج D مرة واحدة رغم إشارة أبوين إليها؛ أما E فهي ممثلة في الرسم، لكن لا يمكن الوصول إليها من A.

def dfs_finish_order(graph):
    seen, finished = set(), []
    for root in graph:
        if root in seen:
            continue
        seen.add(root)
        stack = [(root, iter(graph[root]))]
        while stack:
            vertex, neighbors = stack[-1]
            neighbor = next(neighbors, None)
            if neighbor is None:
                stack.pop()
                finished.append(vertex)
            elif neighbor not in seen:
                seen.add(neighbor)
                stack.append((neighbor, iter(graph[neighbor])))
    return finished

graph = {"A": ["B", "C"], "B": ["D"], "C": ["D"], "D": [], "E": []}
print(dfs_finish_order(graph))

['D', 'B', 'C', 'A', 'E']

يستخدم هذا العقد رؤوسا نصية؛ لذلك يصح استخدام None علامة على نفاد المكرر. ويشكل الجذر الأخير E مكوّنا مستقلا.