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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