Python實現BFS,DFS

BFS:隊

graph = {
    "A" : ["B","C"],
    "B" : ["A","C","D"],
    "C" : ["A","B","D","E"],
    "D" : ["B","C","E","F"],
    "E" : ["C","D"],
    "F" : ["D"]
}
def BFS(graph, s):
    queue = []
    queue.append(s) # 向list添加元素，用append()
    seen = set() # 此處為set, python里set用的是hash table, 搜索時比數組要快。
    seen.add(s) # 向set添加函數，用add()
    while (len(queue) > 0):
        vertex = queue.pop(0)  #提取隊頭
        nodes = graph[vertex]  #獲得隊頭元素的鄰接元素
        for w in nodes:
            if w not in seen:
                queue.append(w) #將沒有遍歷過的子節點入隊
                seen.add(w) #標記好已遍歷
        print("當前出隊的是：",vertex)

BFS(graph, 'A')

 

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 DFS：棧

graph = {
    "A" : ["B","C"],
    "B" : ["A","C","D"],
    "C" : ["A","B","D","E"],
    "D" : ["B","C","E","F"],
    "E" : ["C","D"],
    "F" : ["D"]
}
def DFS(graph, s):
    stack=[]
    stack.append(s) # 向list添加元素，用append()
    seen = set() # 此處為set, python里set用的是hash table, 搜索時比數組要快。
    seen.add(s) # 向set添加函數，用add()
    while (len(stack) > 0):
        vertex = stack.pop() # 彈出最后一個元素
        nodes = graph[vertex]
        for w in nodes:
            if w not in seen:
                stack.append(w)
                seen.add(w)
        print("當前出棧的是",vertex)
DFS(graph, 'A')

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 BFS：求最短路

def BFS2(graph, s):
    queue = []
    queue.append(s)
    seen = set()
    seen.add(s)
    parent = {s:None} #記錄一下父子節點這樣方便求最短路
    while (len(queue) > 0):
        vertex = queue.pop(0)
        nodes = graph[vertex]
        for w in nodes:
            if w not in seen:
                queue.append(w)
                seen.add(w)
                parent[w] = vertex
        print("當前出隊的是：",vertex)
    return parent

parent = BFS2(graph, 'A')
print("父子表：")
for son in parent:
    print(parent[son],son)
print('F->A的最短路：')
start = 'F'
while start:
    print(start,end='->')
    start= parent[start]
print('EDN')

 

 

 

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謝謝燈神的講解，豁然開朗啊喂 ：https://www.bilibili.com/video/av25763384/?spm_id_from=333.788.b_7265636f5f6c697374.2