一、题意理解

给定两棵树T1和T2。如果T1可以通过若干次左右孩子互换就变成T2，则我们称两棵树是“同构的”。现给定两棵树，请你判断它们是否是同构的。

输入格式：输入给出2棵二叉树的信息：

先在一行中给出该树的结点树，随后N行

第i行对应编号第i个结点，给出该结点中存储的字母、其左孩子结点的编号、右孩子结点的编号

如果孩子结点为空，则在相应位置给出“-”

如下图所示，有多种表示的方式，我们列出以下两种：

二、求解思路

搜到一篇也是讲这个的，但是那篇并没有完全用到单向链表的方法，所以研究了一下，写了一个是完全用单向链表的方法：

其实应该有更优雅的删除整个单向列表的方法，比如头设为none，可能会改进下？

# python语言实现L1 = list(map(int, input().split()))L2 = list(map(int, input().split()))# 节点class Node:  def __init__(self, coef, exp):    self.coef = coef    self.exp = exp    self.next = None# 单链表class List:  def __init__(self, node=None):    self.__head = node  # 为了访问私有类  def gethead(self):    return self.__head  def travel(self):    cur1 = self.__head    cur2 = self.__head    if cur1.next != None:      cur1 = cur1.next    else:      print(cur2.coef, cur2.exp, end="")      return    while cur1.next != None:      print(cur2.coef, cur2.exp, end=" ")      cur1 = cur1.next      cur2 = cur2.next    print(cur2.coef, cur2.exp, end=" ")    cur2 = cur2.next    print(cur2.coef, cur2.exp, end="")  # add item in the tail  def append(self, coef, exp):    node = Node(coef, exp)    if self.__head == None:      self.__head = node    else:      cur = self.__head      while cur.next != None:        cur = cur.next      cur.next = nodedef addl(l1, l2):  p1 = l1.gethead()  p2 = l2.gethead()  l3 = List()  while (p1 is not None) & (p2 is not None):    if (p1.exp > p2.exp):      l3.append(p1.coef, p1.exp)      p1 = p1.next    elif (p1.exp < p2.exp):      l3.append(p2.coef, p2.exp)      p2 = p2.next    else:      if (p1.coef + p2.coef == 0):        p1 = p1.next        p2 = p2.next      else:        l3.append(p2.coef + p1.coef, p1.exp)        p2 = p2.next        p1 = p1.next  while p1 is not None:    l3.append(p1.coef, p1.exp)    p1 = p1.next  while p2 is not None:    l3.append(p2.coef, p2.exp)    p2 = p2.next  if l3.gethead() == None:    l3.append(0, 0)  return l3def mull(l1, l2):  p1 = l1.gethead()  p2 = l2.gethead()  l3 = List()  l4 = List()  if (p1 is not None) & (p2 is not None):    while p1 is not None:      while p2 is not None:        l4.append(p1.coef * p2.coef, p1.exp + p2.exp)        p2 = p2.next      l3 = addl(l3, l4)      l4 = List()      p2 = l2.gethead()      p1 = p1.next  else:    l3.append(0, 0)  return l3def L2l(L):  l = List()  L.pop(0)  for i in range(0, len(L), 2):    l.append(L[i], L[i + 1])  return ll1 = L2l(L1)l2 = L2l(L2)l3 = List()l3 = mull(l1, l2)l3.travel()print("")l3 = List()l3 = addl(l1, l2)l3.travel()

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