2022更新国家开放大学电大【数据结构】网络课综合应用题题库及答案

国家开放大学电大【数据构造】网络课综合应用题题库及答案综合应用题题目32(1)以3,4,5,89,作为叶结点的权,构造一梆哈夫曼树.该快J的帝权路径长度为B64 B.65C.62D.66(2)权重为3的叶结卓的哈夫曼循码为CtV.010B.DTD1€.000 D.0111题目33(1)以234,7.8,9作为0简点的权枸『桦昭夫蔓枸岫的希权路径长度为日At66B:80Gr62D:8712＞权重值为4的叶结点的哈夫曼编码为C=力.A.0001?B;1110C.001D.110题目34已知某二叉树的后序遍历序列是debca,中序遍历序列是dbeac,该二叉树的根结点是bSA®B.cC.bD.a先序遍历序列是CV/■A.毛B:.x.a.b.,d.eC;a,b,d,e,eD.a.c,b.dte,题目35(1)已知某二叉树的先序遍历序列是Mcdb,中序颠序列是eadcb,该二叉树的根结点是D€A.:e:Bc D.aA.e.d.D.c.a0.c^.b.^.eCa.b.d.e.cD.a.c.b.d.e.题目36

(1)以结定权重值5,6,17, 25,3D,为叶结点,建立一棵哈夫曼树,该树的中序遍历序列为B§/18.43,2518,43,2518,43,2518.43,2518,43,2518,43,2525,43B5,lb6t28,17,58,30,101.C.5,11.6>28,101,58.30.17.D.5,lb6,28,17,5&30,IDb(2)权.重值为6的叶结点的哈夫曼为Dt/A.1DD1B,D11 C.001 D.00D1题目23•m:(1)设查找表为("0」"4,23,27,29,55,68),画出对上述直噂进展折半直找所对应的判定树,为了成功查用fell需要依次与元素C#V进展•m:A.23.10.1.14 B.23.29.27,14C.23.10.11.14D.23.29.55,14(2)在等概率条件下.成功查找的平均比较次数为A24/9B.25/9 C.3D.25题目24(1)一组记录的关g序列为(47.80,57,39,41.46),利用坷非序的75$也Z的初始堆为B€/(iOS素秘-J盘,祸捌)•A.39.41.57.80,47.46JB.39.41.46.80,47.5739,47,46.80,41,57 D.39.41.57.80,46,47⑵输出堆顶元表后,调枝后的堆为A=y.t♦ ••,!・A.41,47,46.80.57C41,57.80,47,46A.41,47,46.80.57C41,57.80,47,4641.57.46,80,47D.41.80.46,47,57题目25⑴对关谩字序列(56,51.71,54,46JOG).利用快速排序,以第f关键字为分割元素;经过T欠划分后结果为V:;A.46.51.56.54,71JD6; 0,56.51.54,46.71.106C.46,51.54,56.71,106 D.56,51,46,54,71J06(2)一组记录的关键字序列为(60.47<00.57.39.41.46,30).利用归井排序的方法.经过(2.2)归井的结果序列为D#V.A.(30»57,60.00.47.39,41,46) B.(47.60.57.80.30,39.41,46)C,(41t57.60.80.30.39.47,46) D.(47.57.60.80.30,39,41.46)题目26对关键字序列(36,6946,28,30,74)采用快斡E序,以第f关握字为分割屏,经过一次划分后的矣果序列为9令力A30/20,46,36^69,74 0;28i30,36t46.69,74C28i30.46,36.69,74D.30.28.36,46.69,74用冒泡法对上述序列排序,经两通冒泡的结早序列为A§/.A.36,28,30,46.69,74 B36,46.28.20,69,74.C.38.36,30.46.6974 D.28,36,,30.46,69.74题目27(1)一组记录的关键字序列为05,40.65,43,35「95}写出利用快速排序的方法,以第T•记录为暮函导到的划J分的结果为C/354065453595354065434