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一、題意理解

給定兩棵樹T1和T2。如果T1可以通過若干次左右孩子互換就變成T2，則我們稱兩棵樹是“同構(gòu)的”。現(xiàn)給定兩棵樹，請你判斷它們是否是同構(gòu)的。

輸入格式： 輸入給出2棵二叉樹的信息：

• 先在一行中給出該樹的結(jié)點樹，隨后N行

• 第i行對應(yīng)編號第i個結(jié)點，給出該結(jié)點中存儲的字母、其左孩子結(jié)點的編號、右孩子結(jié)點的編號

• 如果孩子結(jié)點為空，則在相應(yīng)位置給出“-”

如下圖所示，有 多種表示 的方式，我們列出以下兩種：

二、求解思路

搜到一篇也是講這個的，但是那篇并沒有完全用到單向鏈表的方法，所以研究了一下，寫了一個是完全用單向鏈表的方法：

其實應(yīng)該有更優(yōu)雅的刪除整個單向列表的方法，比如頭設(shè)為none，可能會改進(jìn)下？

          
            # python語言實現(xiàn)

L1 = list(map(int, input().split()))
L2 = list(map(int, input().split()))


# 節(jié)點
class Node:
    def __init__(self, coef, exp):
        self.coef = coef
        self.exp = exp
        self.next = None


# 單鏈表
class List:
    def __init__(self, node=None):
        self.__head = node

    # 為了訪問私有類
    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()