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568 lines
23 KiB
Markdown
568 lines
23 KiB
Markdown
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---
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title: 『数据结构』树
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date: 2018-7-11 18:56
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categories: 算法与数据结构
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tags: [数据结构,树]
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keywords:
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mathjax: true
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description:
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---
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<!-- TOC -->
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- [1. 概念](#1-概念)
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- [2. 二叉查找树](#2-二叉查找树)
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- [2.1. 随机构造的二叉查找树](#21-随机构造的二叉查找树)
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- [2.2. 平均结点深度](#22-平均结点深度)
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- [2.3. 不同的二叉树数目(Catalan num)](#23-不同的二叉树数目catalan-num)
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- [2.4. 好括号列](#24-好括号列)
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- [3. 基数树(radixTree)](#3-基数树radixtree)
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- [4. 字典树(trie)](#4-字典树trie)
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- [4.1. AC 自动机](#41-ac-自动机)
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- [5. 平衡二叉树](#5-平衡二叉树)
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- [5.1. AVL Tree](#51-avl-tree)
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- [5.2. splayTree](#52-splaytree)
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- [5.2.1. Zig-step](#521-zig-step)
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- [5.2.2. Zig-zig step](#522-zig-zig-step)
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- [5.2.3. Zig-zag step](#523-zig-zag-step)
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- [5.3. read-black Tree](#53-read-black-tree)
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- [6. 总结](#6-总结)
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- [7. 附代码](#7-附代码)
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- [7.1. 二叉树(binaryTree)](#71-二叉树binarytree)
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- [7.2. 前缀树(Trie)](#72-前缀树trie)
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- [7.3. 赢者树(winnerTree)](#73-赢者树winnertree)
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- [7.4. 左斜堆](#74-左斜堆)
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<!-- /TOC -->
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**[github地址](https://github.com/mbinary/algorithm-and-data-struture.git)**
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<a id="markdown-1-概念" name="1-概念"></a>
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# 1. 概念
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* 双亲
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* 左右孩子
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* 左右子树
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* 森林
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* 结点,叶子,边,路径
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* 高度 h
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* 遍历(前中后层)
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* 结点数 n
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<a id="markdown-2-二叉查找树" name="2-二叉查找树"></a>
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# 2. 二叉查找树
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又名排序二叉树,对于每个结点, 如果有,其左孩子不大于它,右孩子不小于它
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通过前序遍历或者后序遍历就可以得到有序序列(升序,降序)
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常用三种操作, 插入,删除,查找,时间复杂度是 $O(h)$
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h是树高, 但是由于插入,删除而导致树不平衡, 即可能 $h\geqslant \lfloor logn \rfloor$
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<a id="markdown-21-随机构造的二叉查找树" name="21-随机构造的二叉查找树"></a>
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## 2.1. 随机构造的二叉查找树
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下面可以证明,随机构造,即输入序列有 $n!$中, 每种概率相同的情况下, 期望的树高 $h=O(logn)$
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(直接搬运算法导论上面的啦>_<)
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<a id="markdown-22-平均结点深度" name="22-平均结点深度"></a>
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## 2.2. 平均结点深度
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一个较 上面定理 弱的结论:
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>一棵随机构造的二叉查找树,n 个结点的平均深度为 $O(logn)$
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类似 RANDOMIZED-QUICKSORT 的证明过程, 因为快排 递归的过程就是一个递归 二叉树.
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随机选择枢纽元就相当于这里的某个子树的根结点 在所有结点的大小随机排名, 如 i. 然后根结点将剩下的结点划分为左子树(i-1)个结点, 右子树(n-i)个结点.
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<a id="markdown-23-不同的二叉树数目catalan-num" name="23-不同的二叉树数目catalan-num"></a>
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## 2.3. 不同的二叉树数目(Catalan num)
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给定$\{1,2,\ldots,n\}$,组成二叉查找树的数目.
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由上面的证明过程, 可以容易地分析得出, 任选第 i 个数作为根, 由于二叉查找树的性质, 其左子树
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应该有 i-1个结点, 右子树有 n-i个结点.
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如果记 n 个结点 的二叉查找树的数目为$b_n$
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则有递推公式
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$$
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b_n=\begin{cases}
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1 &n=0 \\
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\sum_{i=1}^{n}b_{i-1}b_{n-i} & n\geqslant 1
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\end{cases}
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$$
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然后我们来看`<<算法导论>>`(p162,思考题12-4)上怎么求的吧( •̀ ω •́ )y
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设生成函数
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$$B(x)=\sum_{n=0}^{\infty}b_n x^n$$
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下面证明$B(x)=xB(x)^2+1\quad(\#)$
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易得$$xB(x)^2=\sum_{i=1}^{\infty}\sum_{n=i}^{\infty}b_{i-1}b_{n-i}x^n$$
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对比$B(x), xB(x)^2+1$的 x 的各次系数,分别是 $b_k,a_{k}$
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当 k=0, $a_k=1=b_k$
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当 k>0
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$$a_{k} = \sum_{i=1}^{k}b_{i-1}b_{k-i} = b_k$$
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所以$B(x)=xB(x)^2+1\quad(\#)$
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由此解得
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$$B(x)=\frac{1-\sqrt{1-4x} }{2x}$$
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在点 x=0 处,
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用泰勒公式得
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$$
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\begin{aligned}
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\lim_{x\to 0}\sqrt{1-4x}&=1+\sum_{n=1}^{\infty}C_n^{\frac{1}{2}}{(-4)}^nx^n \\
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&=1+\sum_{n=1}^{\infty}\frac{(2n-3)!!{(-4x)}^n}{n!}
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\end{aligned}
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$$
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所以对应系数
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$$
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\begin{aligned}
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b_n&=\frac{1}{2}\frac{4^{n+1}(2n-1)!!}{2^{n+1}n!} \\
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&=\frac{C_{2n}^{n}}{n+1}
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\end{aligned}
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$$
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这个数叫做 `Catalan 数`
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<a id="markdown-24-好括号列" name="24-好括号列"></a>
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## 2.4. 好括号列
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王树禾的`<<图论>>`(p42)上用另外的方法给出Catalan数, 并求出n结点 二叉查找数的个数
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首先定义好括号列,有:
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* 空列,即没有括号叫做好括号列
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* 若A,B都是好括号列, 则串联后 AB是好括号列
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* 若A是好括号列, 则 (A)是好括号列
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>充要条件: 好括号列 <=> 左右括号数相等, 且从左向右看, 看到的右括号数不超过左括号数
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>定理: 由 n个左括号,n个右括号组成的好括号列个数为$c(n)=\frac{C_{2n}^{n}}{n+1}$
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证明:
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由 n左n右组成的括号列有 $\frac{2n}{n!n!}=C_{2n}^{n}$个.
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设括号列$a_1a_2\ldots a_{2n}$为坏括号列,
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由充要条件, 存在最小的 j, 使得$a_1a_2\ldots a_{j}$中右括号比左括号多一个,
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由于是最小的 j, 所以 $a_j$为右括号, $a_{j+1}$为右括号
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把$a_{j+1}a_{j+2}\ldots a_{2n}$中的左括号变为右括号, 右变左,记为$\bar a_{j+1}\bar a_{j+2}\ldots \bar a_{2n}$
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则括号列$a_1a_2\ldots a_{j}\bar a_{j+1}$为好括号列
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$a_1a_2\ldots a_{j}\bar a_{j+1}\bar a_{j+2}\ldots \bar a_{2n}$可好可坏,且有n-1个右,n+1个左, 共有$\frac{2n}{(n+1)!(n-1)!}=C_{2n}^{n+1}$个.
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所以坏括号列$a_1a_2\ldots a_{2n}$ 与括号列 $a_1a_2\ldots a_{j}\bar a_{j+1}\bar a_{j+2}\ldots \bar a_{2n}$, 有$\frac{2n}{(n+1)!(n-1)!}=C_{2n}^{n+1}$个
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那么好括号列有
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$$
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c(n)=C_{2n}^{n} - C_{2n}^{n+1} =\frac{C_{2n}^{n}}{n+1}
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$$
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>推论: n个字符,进栈出栈(出栈可以在栈不为空的时候随时进行), 则出栈序列有 c(n)种
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这种先入后出的情形都是这样
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<a id="markdown-3-基数树radixtree" name="3-基数树radixtree"></a>
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# 3. 基数树(radixTree)
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<a id="markdown-4-字典树trie" name="4-字典树trie"></a>
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# 4. 字典树(trie)
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又叫`前缀树`(preifx tree).适用于储存有公共前缀的字符串集合. 如果直接储存, 而很多字符串有公共前缀, 会浪费掉存储空间.
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字典树可以看成是基数树的变形, 每个结点可以有多个孩子, 每个结点存储的是一个字符, 从根沿着结点走到一个结点,走过的路径形成字符序列, 如果有合适的单词就可以输出.
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当然,也可以同理得出后缀树
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<a id="markdown-41-ac-自动机" name="41-ac-自动机"></a>
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## 4.1. AC 自动机
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Aho-Corasick automation,是在字典树上添加匹配失败边(失配指针), 实现字符串搜索匹配的算法.
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图中蓝色结点 表示存在字符串, 灰色表示不存在.
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黑色边是父亲到子结点的边, 蓝色边就是`失配指针`.
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蓝色边(终点称为起点的后缀结点): 连接字符串**终点**到在**图中存在的**, **最长**严格后缀的结点. 如 caa 的严格后缀为 aa,a, 空. 而在图中存在, 且最长的是字符串 a, 则连接到这个字符串的终点 a.
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绿色边(字典后缀结点): 终点是起点经过蓝色有向边到达的第一个蓝色结点.
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下面摘自 `wiki`
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>在每一步中，算法先查找当前节点的 “孩子节点”，如果没有找到匹配，查找它的后缀节点(suffix) 的孩子，如果仍然没有，接着查找后缀节点的后缀节点的孩子, 如此循环, 直到根结点，如果到达根节点仍没有找到匹配则结束。
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>
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>当算法查找到一个节点，则输出所有结束在当前位置的字典项。输出步骤为首先找到该节点的字典后缀，然后用递归的方式一直执行到节点没有字典前缀为止。同时，如果该节点为一个字典节点，则输出该节点本身。
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>
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>输入 abccab 后算法的执行步骤如下：
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>
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在每一步中，算法先查找当前节点的 “孩子节点”，如果没有找到匹配，查找它的后缀节点(suffix) 的孩子，如果仍然没有，接着查找后缀节点的后缀节点的孩子, 如此循环, 直到根结点，如果到达根节点仍没有找到匹配则结束。
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当算法查找到一个节点，则输出所有结束在当前位置的字典项。输出步骤为首先找到该节点的字典后缀，然后用递归的方式一直执行到节点没有字典前缀为止。同时，如果该节点为一个字典节点，则输出该节点本身。
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输入 abccab 后算法的执行步骤如下：
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输入字符串的分析 abccab
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节点 剩余字符串 输出：结束位置 过渡 输出
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() abccab 从根开始
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(a) bccab a:1 ()→子节点 (a) 当前节点
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(ab) ccab ab:2 (a)→子节点 (ab) 当前节点
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(bc) cab bc:3, c:3 (ab)→后缀节点 (b)→子节点 (bc) 当前节点，字典后缀
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(c) ab c:4 (bc)→后缀节点 (c)→后缀节点 ()→子节点 (c) 当前节点
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(ca) b a:5 (c)→(ca) 字典后缀
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(ab) ab:6 (ca)→后缀节点 (a)→子节点 (ab) 当前节点
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<a id="markdown-5-平衡二叉树" name="5-平衡二叉树"></a>
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# 5. 平衡二叉树
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上面的二叉查找树不平衡,即经过多次插入,删除后, 其高度变化大, 不能保持$\Theta(n)$的性能
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而平衡二叉树就能.
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平衡二叉树都是经过一些旋转操作, 使左右子树的结点高度相差不大,达到平衡
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有如下几种
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<a id="markdown-51-avl-tree" name="51-avl-tree"></a>
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## 5.1. AVL Tree
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`平衡因子`: 右子树高度 - 左子树高度
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定义: 每个结点的平衡因子属于{0,-1,1}
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<a id="markdown-52-splaytree" name="52-splaytree"></a>
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## 5.2. splayTree
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伸展树, 它的特点是每次将访问的结点通过旋转旋转到根结点.
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其实它并不平衡. 但是插入,查找,删除操作 的平摊时间是$O(logn)$
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有三种旋转,下面都是将访问过的 x 旋转到 根部
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<a id="markdown-521-zig-step" name="521-zig-step"></a>
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### 5.2.1. Zig-step
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<a id="markdown-522-zig-zig-step" name="522-zig-zig-step"></a>
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### 5.2.2. Zig-zig step
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<a id="markdown-523-zig-zag-step" name="523-zig-zag-step"></a>
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### 5.2.3. Zig-zag step
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<a id="markdown-53-read-black-tree" name="53-read-black-tree"></a>
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## 5.3. read-black Tree
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同样是平衡的二叉树, 以后单独写一篇关于红黑树的.
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<a id="markdown-6-总结" name="6-总结"></a>
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# 6. 总结
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还有很多有趣的树结构,
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比如斜堆, 竞赛树(赢者树,输者树,线段树, 索引树,B树, fingerTree(不知道是不是译为手指树233)...
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这里就不详细介绍了, 如果以后有时间,可能挑几个单独写一篇文章
|
||
|
|
|
||
|
|
|
||
|
|
<a id="markdown-7-附代码" name="7-附代码"></a>
|
||
|
|
# 7. 附代码
|
||
|
|
<a id="markdown-71-二叉树binarytree" name="71-二叉树binarytree"></a>
|
||
|
|
## 7.1. 二叉树(binaryTree)
|
||
|
|
```python
|
||
|
|
from functools import total_ordering
|
||
|
|
|
||
|
|
@total_ordering
|
||
|
|
class node:
|
||
|
|
def __init__(self,val,left=None,right=None,freq = 1):
|
||
|
|
self.val=val
|
||
|
|
self.left=left
|
||
|
|
self.right=right
|
||
|
|
self.freq = freq
|
||
|
|
def __lt__(self,nd):
|
||
|
|
return self.val<nd.val
|
||
|
|
def __eq__(self,nd):
|
||
|
|
return self.val==nd.val
|
||
|
|
def __repr__(self):
|
||
|
|
return 'node({})'.format(self.val)
|
||
|
|
|
||
|
|
class binaryTree:
|
||
|
|
def __init__(self):
|
||
|
|
self.root=None
|
||
|
|
def add(self,val):
|
||
|
|
def _add(nd,newNode):
|
||
|
|
if nd<newNode:
|
||
|
|
if nd.right is None:nd.right = newNode
|
||
|
|
else:_add(nd.right,newNode)
|
||
|
|
elif nd>newNode:
|
||
|
|
if nd.left is None:nd.left = newNode
|
||
|
|
else : _add(nd.left,newNode)
|
||
|
|
else:nd.freq +=1
|
||
|
|
_add(self.root,node(val))
|
||
|
|
def find(self,val):
|
||
|
|
prt= self._findPrt(self.root,node(val),None)
|
||
|
|
if prt.left and prt.left.val==val:
|
||
|
|
return prt.left
|
||
|
|
elif prt.right and prt.right.val==val:return prt.right
|
||
|
|
else :return None
|
||
|
|
def _findPrt(self,nd,tgt,prt):
|
||
|
|
if nd==tgt or nd is None:return prt
|
||
|
|
elif nd<tgt:return self._findPrt(nd.right,tgt,nd)
|
||
|
|
else:return self._findPrt(nd.left,tgt,nd)
|
||
|
|
def delete(self,val):
|
||
|
|
prt= self._findPrt(self.root,node(val),None)
|
||
|
|
if prt.left and prt.left.val==val:
|
||
|
|
l=prt.left
|
||
|
|
if l.left is None:prt.left = l.right
|
||
|
|
elif l.right is None : prt.left = l.left
|
||
|
|
else:
|
||
|
|
nd = l.left
|
||
|
|
while nd.right is not None:nd = nd.right
|
||
|
|
nd.right = l.right
|
||
|
|
prt.left = l.left
|
||
|
|
elif prt.right and prt.right.val==val:
|
||
|
|
r=prt.right
|
||
|
|
if r.right is None:prt.right = r.right
|
||
|
|
elif r.right is None : prt.right = r.left
|
||
|
|
else:
|
||
|
|
nd = r.left
|
||
|
|
while nd.right is not None:nd = nd.right
|
||
|
|
nd.right = r.right
|
||
|
|
prt.left = r.left
|
||
|
|
|
||
|
|
def preOrder(self):
|
||
|
|
def _p(nd):
|
||
|
|
if nd is not None:
|
||
|
|
print(nd)
|
||
|
|
_p(nd.left)
|
||
|
|
_p(nd.right)
|
||
|
|
_p(self.root)
|
||
|
|
```
|
||
|
|
<a id="markdown-72-前缀树trie" name="72-前缀树trie"></a>
|
||
|
|
## 7.2. 前缀树(Trie)
|
||
|
|
```python
|
||
|
|
class node:
|
||
|
|
def __init__(self,val = None):
|
||
|
|
self.val = val
|
||
|
|
self.isKey = False
|
||
|
|
self.children = {}
|
||
|
|
def __getitem__(self,i):
|
||
|
|
return self.children[i]
|
||
|
|
def __iter__(self):
|
||
|
|
return iter(self.children.keys())
|
||
|
|
def __setitem__(self,i,x):
|
||
|
|
self.children[i] = x
|
||
|
|
def __bool__(self):
|
||
|
|
return self.children!={}
|
||
|
|
def __str__(self):
|
||
|
|
return 'val: '+str(self.val)+'\nchildren: '+' '.join(self.children.keys())
|
||
|
|
def __repr__(self):
|
||
|
|
return str(self)
|
||
|
|
|
||
|
|
class Trie(object):
|
||
|
|
|
||
|
|
def __init__(self):
|
||
|
|
self.root=node('')
|
||
|
|
self.dic ={'insert':self.insert,'startsWith':self.startsWith,'search':self.search}
|
||
|
|
|
||
|
|
def insert(self, word):
|
||
|
|
"""
|
||
|
|
Inserts a word into the trie.
|
||
|
|
:type word: str
|
||
|
|
:rtype: void
|
||
|
|
"""
|
||
|
|
if not word:return
|
||
|
|
nd = self.root
|
||
|
|
for i in word:
|
||
|
|
if i in nd:
|
||
|
|
nd = nd[i]
|
||
|
|
else:
|
||
|
|
newNode= node(i)
|
||
|
|
nd[i] = newNode
|
||
|
|
nd = newNode
|
||
|
|
else:nd.isKey = True
|
||
|
|
def search(self, word,matchAll='.'):
|
||
|
|
"""support matchall function eg, 'p.d' matchs 'pad' , 'pid'
|
||
|
|
"""
|
||
|
|
self.matchAll = '.'
|
||
|
|
return self._search(self.root,word)
|
||
|
|
def _search(self,nd,word):
|
||
|
|
for idx,i in enumerate(word):
|
||
|
|
if i==self.matchAll :
|
||
|
|
for j in nd:
|
||
|
|
bl =self._search(nd[j],word[idx+1:])
|
||
|
|
if bl:return True
|
||
|
|
else:return False
|
||
|
|
if i in nd:
|
||
|
|
nd = nd[i]
|
||
|
|
else:return False
|
||
|
|
else:return nd.isKey
|
||
|
|
def startsWith(self, prefix):
|
||
|
|
"""
|
||
|
|
Returns if there is any word in the trie that starts with the given prefix.
|
||
|
|
:type prefix: str
|
||
|
|
:rtype: bool
|
||
|
|
"""
|
||
|
|
nd = self.root
|
||
|
|
for i in prefix:
|
||
|
|
if i in nd:
|
||
|
|
nd= nd[i]
|
||
|
|
else:return False
|
||
|
|
return True
|
||
|
|
def display(self):
|
||
|
|
print('preOrderTraverse data of the Trie')
|
||
|
|
self.preOrder(self.root,'')
|
||
|
|
def preOrder(self,root,s):
|
||
|
|
s=s+root.val
|
||
|
|
if root.isKey:
|
||
|
|
print(s)
|
||
|
|
for i in root:
|
||
|
|
self.preOrder(root[i],s)
|
||
|
|
```
|
||
|
|
<a id="markdown-73-赢者树winnertree" name="73-赢者树winnertree"></a>
|
||
|
|
## 7.3. 赢者树(winnerTree)
|
||
|
|
```python
|
||
|
|
class winnerTree:
|
||
|
|
'''if i<lowExt p = (i+offset)//2
|
||
|
|
else p = (i+n-1-lowExt)//2
|
||
|
|
offset is a num 2^k-1 just bigger than n
|
||
|
|
p is the index of tree
|
||
|
|
i is the index of players
|
||
|
|
lowExt is the double node num of the lowest layer of the tree
|
||
|
|
'''
|
||
|
|
def __init__(self,players,reverse=False):
|
||
|
|
self.n=len(players)
|
||
|
|
self.tree = [0]*self.n
|
||
|
|
players.insert(0,0)
|
||
|
|
self.players=players
|
||
|
|
self.reverse=reverse
|
||
|
|
self.getNum()
|
||
|
|
self.initTree(1)
|
||
|
|
def getNum(self):
|
||
|
|
i=1
|
||
|
|
while 2*i< self.n:i=i*2
|
||
|
|
if 2*i ==self. n:
|
||
|
|
self.lowExt=0
|
||
|
|
self.s = 2*i-1
|
||
|
|
else:
|
||
|
|
self.lowExt = (self.n-i)*2
|
||
|
|
self.s = i-1
|
||
|
|
self.offset = 2*i-1
|
||
|
|
def treeToArray(self,p):
|
||
|
|
return 2*p-self.offset if p>self.s else 2*p+self.lowExt-self.n+1
|
||
|
|
def arrayToTree(self,i):
|
||
|
|
return (i+self.offset)//2 if i<=self.lowExt else (i-self.lowExt+ self.n-1)//2
|
||
|
|
def win(self,a,b):
|
||
|
|
return a<b if self.reverse else a>b
|
||
|
|
def initTree(self,p):
|
||
|
|
if p>=self.n:
|
||
|
|
delta = p%2 #!!! good job notice delta mark the lchild or rchlid
|
||
|
|
return self.players[self.treeToArray(p//2)+delta]
|
||
|
|
l = self.initTree(2*p)
|
||
|
|
r = self.initTree(2*p+1)
|
||
|
|
self.tree[p] = l if self.win(l,r) else r
|
||
|
|
return self.tree[p]
|
||
|
|
def winner(self):
|
||
|
|
idx = 1
|
||
|
|
while 2*idx<self.n:
|
||
|
|
idx = 2*idx if self.tree[2*idx] == self.tree[idx] else idx*2+1
|
||
|
|
num = self.treeToArray(idx)
|
||
|
|
num = num+1 if self.players[num] !=self.tree[1] else num
|
||
|
|
return self.tree[1],num
|
||
|
|
def getOppo(self,i,x,p):
|
||
|
|
oppo=None
|
||
|
|
if 2*p<self.n:oppo=self.tree[2*p]
|
||
|
|
elif i<=self.lowExt:oppo=self.players[i-1+i%2*2]
|
||
|
|
else:
|
||
|
|
lpl= self.players[2*p+self.lowExt-self.n+1]
|
||
|
|
oppo = lpl if lpl!=x else self.players[2*p+self.lowExt-self.n+2]
|
||
|
|
return oppo
|
||
|
|
def update(self,i,x):
|
||
|
|
''' i is 1-indexed which is the num of player
|
||
|
|
and x is the new val of the player '''
|
||
|
|
self.players[i]=x
|
||
|
|
p = self.arrayToTree(i)
|
||
|
|
oppo =self.getOppo(i,x,p)
|
||
|
|
self.tree[p] = x if self.win(x,oppo) else oppo
|
||
|
|
p=p//2
|
||
|
|
while p:
|
||
|
|
l = self.tree[p*2]
|
||
|
|
r = None
|
||
|
|
if 2*p+1<self.n:r=self.tree[p*2+1] #notice this !!!
|
||
|
|
else:r = self.players[2*p+self.lowExt-self.n+1]
|
||
|
|
self.tree[p] = l if self.win(l,r) else r
|
||
|
|
p=p//2
|
||
|
|
```
|
||
|
|
<a id="markdown-74-左斜堆" name="74-左斜堆"></a>
|
||
|
|
## 7.4. 左斜堆
|
||
|
|
```python
|
||
|
|
from functools import total_ordering
|
||
|
|
@total_ordering
|
||
|
|
|
||
|
|
class node:
|
||
|
|
def __init__(self,val,freq=1,s=1,left=None,right=None):
|
||
|
|
self.val=val
|
||
|
|
self.freq=freq
|
||
|
|
self.s=s
|
||
|
|
if left is None or right is None:
|
||
|
|
self.left = left if left is not None else right
|
||
|
|
self.right =None
|
||
|
|
else:
|
||
|
|
if left.s<right.s:
|
||
|
|
left,right =right, left
|
||
|
|
self.left=left
|
||
|
|
self.right=right
|
||
|
|
self.s+=self.right.s
|
||
|
|
def __eq__(self,nd):
|
||
|
|
return self.val==nd.val
|
||
|
|
def __lt__(self,nd):
|
||
|
|
return self.val<nd.val
|
||
|
|
def __repr__(self):
|
||
|
|
return 'node(val=%d,freq=%d,s=%d)'%(self.val,self.freq,self.s)
|
||
|
|
|
||
|
|
class leftHeap:
|
||
|
|
def __init__(self,root=None):
|
||
|
|
self.root=root
|
||
|
|
def __bool__(self):
|
||
|
|
return self.root is not None
|
||
|
|
@staticmethod
|
||
|
|
def _merge(root,t): #-> int
|
||
|
|
if root is None:return t
|
||
|
|
if t is None:return root
|
||
|
|
if root<t:
|
||
|
|
root,t=t,root
|
||
|
|
root.right = leftHeap._merge(root.right,t)
|
||
|
|
if root.left is None or root.right is None:
|
||
|
|
root.s=1
|
||
|
|
if root.left is None:
|
||
|
|
root.left,root.right = root.right,None
|
||
|
|
else:
|
||
|
|
if root.left.s<root.right.s:
|
||
|
|
root.left,root.right = root.right,root.left
|
||
|
|
root.s = root.right.s+1
|
||
|
|
return root
|
||
|
|
def insert(self,nd):
|
||
|
|
if not isinstance(nd,node):nd = node(nd)
|
||
|
|
if self.root is None:
|
||
|
|
self.root=nd
|
||
|
|
return
|
||
|
|
if self.root==nd:
|
||
|
|
self.root.freq+=1
|
||
|
|
return
|
||
|
|
prt =self. _findPrt(self.root,nd,None)
|
||
|
|
if prt is None:
|
||
|
|
self.root=leftHeap._merge(self.root,nd)
|
||
|
|
else :
|
||
|
|
if prt.left==nd:
|
||
|
|
prt.left.freq+=1
|
||
|
|
else:prt.right.freq+=1
|
||
|
|
def remove(self,nd):
|
||
|
|
if not isinstance(nd,node):nd = node(nd)
|
||
|
|
if self.root==nd:
|
||
|
|
self.root=leftHeap._merge(self.root.left,self.root.right)
|
||
|
|
else:
|
||
|
|
prt = self._findPrt(self.root,nd,None)
|
||
|
|
if prt is not None:
|
||
|
|
if prt.left==nd:
|
||
|
|
prt.left=leftHeap._merge(prt.left.left,prt.left.right)
|
||
|
|
else:
|
||
|
|
prt.right=leftHeap._merge(prt.right.left,prt.right.right)
|
||
|
|
def find(self,nd):
|
||
|
|
if not isinstance(nd,node):nd = node(nd)
|
||
|
|
prt = self._findPrt(self.root,nd,self.root)
|
||
|
|
if prt is None or prt==nd:return prt
|
||
|
|
elif prt.left==nd:return prt.left
|
||
|
|
else:return prt.right
|
||
|
|
def _findPrt(self,root,nd,parent):
|
||
|
|
if not isinstance(nd,node):nd = node(nd)
|
||
|
|
if root is None or root<nd:return None
|
||
|
|
if root==nd:return parent
|
||
|
|
l=self._findPrt(root.left,nd,root)
|
||
|
|
return l if l is not None else self._findPrt(root.right,nd,root)
|
||
|
|
def getTop(self):
|
||
|
|
return self.root
|
||
|
|
def pop(self):
|
||
|
|
nd = self.root
|
||
|
|
self.remove(self.root.val)
|
||
|
|
return nd
|
||
|
|
def levelTraverse(self):
|
||
|
|
li = [(self.root,0)]
|
||
|
|
cur=0
|
||
|
|
while li:
|
||
|
|
nd,lv = li.pop(0)
|
||
|
|
if cur<lv:
|
||
|
|
cur=lv
|
||
|
|
print()
|
||
|
|
print(nd,end=' ')
|
||
|
|
else:print(nd,end=' ')
|
||
|
|
if nd.left is not None:li.append((nd.left,lv+1))
|
||
|
|
if nd.right is not None:li.append((nd.right,lv+1))
|
||
|
|
```
|