目录

• 一、题意理解
• 二、求解思路

更新、更全的《数据结构与算法》的更新网站，更有python、go、人工智能教学等着你：<https://www.cnblogs.com/nickchen121/p/11407287.html

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 = node


def 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 l3


def 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 l3


def L2l(L):
    l = List()
    L.pop(0)
    for i in range(0, len(L), 2):
        l.append(L[i], L[i + 1])
    return l


l1 = L2l(L1)
l2 = L2l(L2)
l3 = List()
l3 = mull(l1, l2)
l3.travel()
print("")
l3 = List()
l3 = addl(l1, l2)
l3.travel()

标签：p2,专场,同构,p1,coef,python,next,l3,exp
来源： https://www.cnblogs.com/nickchen121/p/11518875.html