Class で DFS

DFSを毎回自力で実装するのがめんどくさいのでAtcoder のKiの問題をclassで実装したのでメモとして残します

以下　利用したコード

N,Q = map(int,input().split())
Graph = [[]for _ in range(N)]
operater = []
for _ in range(N-1):
    a,b = map(int,input().split())
    Graph[a-1].append(b-1)
    Graph[b-1].append(a-1)
from collections import deque
q = deque()
q.append(0)
seen = [False for _ in range(N)]
parent = [[] for _ in range(N)]#i番目の値はi-1頂点の親のリスト
while q:
    now = q.popleft()#int
    seen[now] = True
    next = Graph[now]#list
    for ver in next:
        if seen[ver] == False:
            q.appendleft(ver)
            parent[ver].append(now)#深さ優先探索で親を求めた
counter = [0 for _ in range(N)]
class DFS:
    def __init__(self,N,Graph,parent,counter):
        self.seen = [False for _ in range(N)]
        self.delete = []
    def children (self,st):#自分を含む子をすべて返す
        self.ans = []
        for i in parent[st]:
            self.seen[i] = True
            self.delete.append(i)
        from collections import deque
        q = deque()
        q.append(st)
        while q:
            now = q.popleft()  # int
            self.seen[now] = True
            next = Graph[now]  # list
            for ver in next:
                if self.seen[ver] == False:
                    q.appendleft(ver)
        for i,j in enumerate(self.seen):
            if j == True and i not in self.delete:
                self.ans.append(i)
        self.seen = [False for _ in range(N)]
        self.delete = []
        return self.ans
    def count(self,p,x):
        self.c = self.children(p)
        for i in self.c:
            counter[i] += x
        return counter
#このクラスでは予め親リストをつくって渡していることに注意。いつか親リストをつくるところもクラスにします。