请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的。
class Node(): def __init__(self,val): self.val = val self.lchild = None self.rchild = None class Tree(): def __init__(self,root): self.root = root def add(self,item): node = Node(item) if not self.root: self.root = node return else: quene = [self.root] while quene: cur_node = quene.pop(0) if not cur_node.lchild: cur_node.lchild = node return else: quene.append(cur_node.lchild) if not cur_node.rchild: cur_node.rchild = node print(222222,node.val) return else: quene.append(cur_node.rchild) # def seek(self): # if not root: # return # #方法1：广度遍历 # llist = [self.root.lchild] # rlist = [self.root.rchild] # lquene = [self.root.lchild] # rquene = [self.root.rchild] # print(rquene) # # 广度遍历左子树 # while lquene: # cur_node = lquene.pop(0) # if cur_node.lchild: # lquene.append(cur_node.lchild) # llist.append(cur_node.lchild) # if cur_node.rchild: # lquene.append(cur_node.rchild) # llist.append(cur_node.rchild) # #广度遍历右子树 # while rquene: # cur_node = rquene.pop(0) # if cur_node.lchild: # rquene.append(cur_node.lchild) # llist.append(cur_node.lchild) # if cur_node.rchild: # rquene.append(cur_node.rchild) # rlist.append(cur_node.rchild) # n1 = len(llist) # n2 = len(rlist) # l1 = [i.val for i in llist] # l2 = [i.val for i in rlist] # print(l1,l2) # if n1 != n2: # return False # for i in range(n1): # if llist[i].val != rlist[i].val: # return False # return True # #方法二：深度遍历 #深度遍历（先序） def seek(self): if not self.root: return def deep_xian(root,a): if root==None: return else: a.append(root) deep_xian(root.lchild,a) deep_xian(root.rchild,a) llist = [] rlist = [] deep_xian(root.lchild,llist) deep_xian(root.rchild,rlist) n1 = len(llist) n2 = len(rlist) l1 = [i.val for i in llist] l2 = [i.val for i in rlist] print(l1,l2) if n1 != n2: return False for i in range(n1): if llist[i].val != rlist[i].val: return False return True # 测试 root = Node(1) etree = Tree(root) etree.add(2) etree.add(3) etree.add(4) etree.add(5) etree.add(6) etree.add(7) etree.add(8) etree.add(9) etree.add(10) print(etree.seek())