10215501439.rr.lesson4.ipynb

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      "11001011.10110011.11001.100101\n"
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    "\"\"\"请编写一个函数，\n",
    "    利用 辗转除2取余法（不要使用int的直接转换） \n",
    "    实现将我们常见的十进制IPv4地址例如203.179.25.37转换为32位长的二进制地址。\n",
    "\"\"\"\n",
    "#成功！\n",
    "def ten22(n):#辗转除2取余法   \n",
    "       #利用的基本对象：列表（列表可以存放一切类型，类似于一个类）\n",
    "        b = []  # 存储余数\n",
    "        while True:  # 一直循环，商为0时利用break退出循环\n",
    "            s = n // 2  # 商\n",
    "            y = n % 2  # 余数\n",
    "            b = b + [y]  # 每一个余数存储到b中（加列表简单粗暴）\n",
    "            if s == 0:\n",
    "                break  # 余数为0时结束循环\n",
    "            n = s\n",
    "        b.reverse()  # 使b中的元素反向排列\n",
    "        res=int(''.join(list(map(str,b))))#将列表中的数字（可能是数字字符串，可能是单个数字字符）转化为int型数字\n",
    "        return res\n",
    "    \n",
    "def convert_ip_address_to_dec(address):\n",
    "    firstlist=[]\n",
    "    for i in range(4):#注意，i取到0,1,2,3\n",
    "        firstlist.append(address.split('.',3)[i])#利用split函数，和显著的分隔符‘。’，3表示一共分为3+1段，【i】表示选取第i段\n",
    "    secondlist=[]\n",
    "    for add in firstlist:\n",
    "        secondlist.append(ten22(int(''.join(list(map(str,add))))))#传入ten22函数的是数字，故转换一下\n",
    "    result=str(secondlist[0])+'.'+str(secondlist[1])+'.'+str(secondlist[2])+'.'+str(secondlist[3])\n",
    "    return result\n",
    "\n",
    "address='203.179.25.37'\n",
    "print(convert_ip_address_to_dec(address))\n",
    "#难题，类型转换"
   ]
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    "\"\"\"\n",
    "请创建一个类模拟实现数据结构中的“队列”，\n",
    "类中应包含入队、出队、取队首/队尾元素等方法，并进行测试。\n",
    "\"\"\"\n",
    "class Quene():\n",
    "    def _init_(self):\n",
    "        self._items = []\n",
    "    def size(self):\n",
    "        return len(self._items)\n",
    "    def isempty(self):\n",
    "        if len(self._items)==0:\n",
    "            return True\n",
    "        else:\n",
    "            return False\n",
    "    def inquene(self,element):\n",
    "        self._items.append(element)\n",
    "    def pop(self):\n",
    "        try:\n",
    "            return self._items.pop(0)\n",
    "        except: \n",
    "            print('ERROR:Quene is empty now!!')\n",
    "    def peek(self):\n",
    "        try:\n",
    "            return self._items[0]\n",
    "        except:\n",
    "            print('ERROR:Quene is empty now!!')\n",
    "\n",
    "q=Quene()\n",
    "#inquene(2)\n",
    "print(q.pop()) # 堆栈目前为 [ ]\n",
    "print(q.pop()) # 错误，堆栈为空\n",
    "\n",
    "print('**********')\n",
    "\n",
    "q.inquene(3.5)  # 堆栈目前为[3.5]\n",
    "q.inquene(2.7)  # 堆栈目前为[3.5, 2.7]\n",
    "print(q.peek())\n",
    "print(q.size())\n",
    "print(q.isempty())"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
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   "source": [
    "\"\"\"\n",
    "请在实验4-4的基础上实现二叉树的层序遍历的函数。\n",
    "提示：需要用到队列\n",
    "\"\"\"\n",
    "class BinaryTree:\n",
    "    def __init__(self,data=None,left=None,right=None):  # 如果创建节点对象时left或right参数为空，则默认该节点没有左或右子树\n",
    "        self.data=data\n",
    "        self.left=left\n",
    "        self.right=right\n",
    "    def breadth_travel(self):\n",
    "        # 层序遍历， 即广度优先遍历BFS，用队列的方式实现\n",
    "        #对于空树，直接返回\n",
    "        if self.root is None:\n",
    "            return\n",
    "        #利用队列， 进行循环的判断，当队列为空时，循环结束\n",
    "        queue = [self.root]\n",
    "        while queue:\n",
    "            cur = queue.pop(0)#取当前队首元素\n",
    "            print(cur.data, end=\" \")#因为print的习惯是输出一个，换一行，为了避免换行，用空格代替\n",
    "            if cur.left_child is not None:\n",
    "                queue.append(cur.left_child)\n",
    "            if cur.right_child is not None:\n",
    "                queue.append(cur.right_child)\n"
   ]
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   "cell_type": "code",
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   "source": [
    "class BinaryTree:  \n",
    "    def __init__(self,data=None,left=None,right=None):  # 如果创建节点对象时left或right参数为空，则默认该节点没有左或右子树\n",
    "        self.data=data\n",
    "        self.left=left\n",
    "        self.right=right\n",
    "        \n",
    "    def leaf(self,root):\n",
    "\n",
    "         if root==None:\n",
    "            return 0\n",
    "\n",
    "         elif root.left==None and root.right==None:#注意是使用and\n",
    "            return 1\n",
    "\n",
    "         else:#否则返回左右子树\n",
    "            return  (self.leaf(root.left)+self.leaf(root.right))   \n",
    "\n",
    "\n",
    "BinaryTree.leaf(tree.root)\n",
    "\n"
   ]
  }
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