課程資源-數(shù)據(jù)結(jié)構(gòu)與算法chapter4.0tree

Chapter4TwokindsofdataLinear:list,stack,queue,Non-linear:tree,4.1AtreeTisacollectionofThecollectioncanbeotherwise,atreeconsistsofadistinguishednoder,calledtheroot,andzeroormorenonempty(sub)treesT1,T2,……,Tk Degreeofanelememts(nodes):thenumberchildrenitDegreeofatree: umofitsLeaf:elementwhosedegreeis0Branch:elementwhosedegreeisnot0thelevelofrootis0 thelevelofanelement=thelevelofitsDepthofa umlevelofitsBinaryDefinition:Abinarytreetisafinite(possiblyempty)collectionofelements.WhenthebinarytreeisnotIthasarootTheremainingelements(ifany)arepartitionedintotwobinarytrees,whicharecalledtheleftandrightsubtreesoft.BinaryTheessentialdifferencesbetweenabinarytreeandatreeare:Eachelementinabinarytreehasexactlysubtrees(oneorbothofthesesubtreesmaybeEachelementinatreecanhaveanynumberofThesubtreesofeachelementinabinarytreeareThatis,wedistinguishbetweentheleftandtheThesubtreesinatreeareBinaryExampleofabinary++*/abcdPropertiesofbinaryProperty1.Thedrawingofeverybinarytreewithnelements(n>0)hasexactlyn-1edges.Property2.Thenumberofelementsatleveliisatmost2i(i>=0).PropertiesofbinaryProperty3.Abinarytreeofheighth,h>=0,hasatleasth+1andatmost2h+1–1elementsinofofproperty

PropertiesofbinaryProperty4.Ifnumberofleavesisn0,andthenumberofthe2degreeelementsisn2,thenn0=n2+1. 度為1的結(jié)點(diǎn)數(shù)是n1個(gè)n=n0+n1+n2n 這里B為分支n0+n1+n2=1*n1+2*n2+n0=n2+PropertiesofbinaryProperty5.Theheightofabinarytreethatcontainsn(n>=0)elementisatmostn-1andatleastproof:Sincetheremustbeatleastoneelementateachlevel,theheightcannotexceedn-1.Fromproperty3,weknown<=2h+1-1,so,h>=log2(n+1)-1,sincehisaninteger,wegetPropertiesofbinaryDefinitionofafullbinarytreeAbinarytreeofheighththatcontainsexactly2h+1-1elementsiscalledafullbinarytree.PropertiesofbinaryDefinitionofacompletebinarySupposewenumbertheelementsinafullbinarytreeofheighthusingthenumber1throughWebeganatlevel0andgodowntolevelh.WithinlevelstheelementsarenumberedlefttoSupposewedeletethekelementsnumbered2h+1-i,1<=i<=k,theresultingbinarytreeiscalledacompletebinarytree.65Propertiesofbinary653Exampleofcompletebinary3123123412412PropertiesofbinaryProperty6.Leti,0<=i<=n-1,bethenumberassignedtoelementofacompletebinarytree.Thefollowingareifi=0,thenthiselementistherootofthebinaryifi>0,thentheparentofthiselementhasbeenassignedthenumber(i-1)/2if2*i+1>=n,thenthiselementhasnoleftchild.leftchildhasbeenassignedthenumber4.3Propertiesofbinaryif2*i+2>=n,thenthiselementhasnorightchild,Otherwiseitsrightchildhasbeenassignedthenumber4.4RepresentationofbinaryFormula-BasedRepresentation(arrayThebinarytreetoberepresentedisregardedasacompletebinarytreewithsomemissingelements.RepresentationofbinaryExampleofacompletebinarytree(array RepresentationofbinaryExampleofacommonbinarytree(array

910113123126 RepresentationofbinaryLinkedrepresentation(alsocalledL-RlinkedThenode Representationofbinary ABCDABCDEFGBDF^

^C^^^C^^E^^G^RepresentationofbinaryA1-A1-B23C--D45E-6F--G--01234564.4RepresentationofbinaryNodeclassforlinkedbinarytreesclassBianryNode{BinaryNode(Objecte){element=e;Left=Right=0;}BinaryNode(Objecte,BinaryNodel,BinaryNoder){element=e;Left=l;Right=r;ObjectBinaryNodeleft; //leftsubtreeBinaryNoderight; //rightsubtree4.5Commonbinarytree datatypebinaryMakeTree(root,left,BreakTree(root,left,CommonbinarytreeTheClassBinarytreetemplate<classT>class{bool{returnboolRoot(T&voidMakeTree(constT&BinaryTree<T>&leftch,BinaryTree<T>&CommonbinarytreevoidBreakTree(T&data,BinaryTree<T>&leftch,BinaryTree<T>&rightch);void{PreOrder(visit,void{InOrder(visit,voidPostOrder{PostOrder(visit,root);}voidLevelOrder(void(*visit)(BinaryNode<T>CommonbinarytreeBinaryNode<T>*voidPreOrder(void(*visit)(BinaryNode<T>voidInOrder(void(*visit)(BinaryNode<T>voidPostOrder(void(*visit)(BinaryNode<T>CommonbinarytreeInthisclass,weemployalinkedrepresentationforbinarytrees.Thefunctionvisitisusedasparametertothetraversalmethods,sothatdifferentoperationscanbeimplementedeasily4.5CommonbinarytreeImplementationofsomememberfunctionsTemplate<classT>voidBinaryTree<T>::MakeTree(constT&data,BinaryTree<T>&leftch, {root=newBinaryNode<T>(data,}4.5Commonbinarytreetemplate<classvoidBinaryTree<T>::BreakTree(T&data,BinaryTree<T>&leftch,BinaryTree<T>&rightch){if(!root)throwBadInput();//treedata=root.element;leftch.root=root.Left;rightch.root=root.Right;deleteroot;}4.5CommonbinarytreeApplicationofclassBinaryTree(Create#include“binary.h”int template<classvoidvoid{}4.6BinaryTreeEachelementisvisitedexactlyV-----表 一個(gè)結(jié)L------表 V的R------表 V的

LVRLRVVRLRVLLevel4.6BinaryTreeExampleofbinarytreeABCDEABCDEFGHILevelorder:4.6BinaryTreePreorder{//preordertraversalof*t. }}BinaryTreeInorder{if(t){}}BinaryTreePostordertraversal{}}4.6BinaryTreeInorderBinaryTreeLevelitisanon-recursivefunctionandaqueueisABABCDEFGHIIHGFEDCBABinaryTreeLeveltemplate<classvoidLevelOrder(BinaryNode<T>*{LinkedQueue<BinaryNode<T>*> //visitif(t→Left)Q.Add(t→Left);}}Inorder,Postordernon-recursiveInordernon-recursiveABABCDEFGHIInordernon-recursivevoidInorder(BinaryNode<T>*{Stack<BinaryNode<T>*>BinaryNode<T>*p=for(;;{1){ p=p->Left;if(!s.IsEmpty({p=s.pop(cout<<p-p=p-}else}}Inorder,Postordernon-recursivePostordernon-recursiveA struct

{BinaryNode<T>*int}Postordernon-recursivevoidPostorder(BinaryNode<T>*{Stack<StkNode<T>>s(10);StkNode<T>Cnode;BinaryNode<T>*p=t;for(;;){1)while{Cnode.ptr=p;p=p->Left;

Cnode.tag=0;}2)Cnode=s.pop( p=while(Cnode.tag== //從 {cout<<p-if(!s.IsEmpty({Cnode=s.pop(); p=Cnode.ptr;}elsereturn;}4)Cnode.tag=1;}

p=p- //從 回建立一棵二叉樹(shù)的方利用MakeTree函利用先序、中序唯一的構(gòu)造一棵二叉先序 中序 G 利用二叉樹(shù)的廣義表表示來(lái)構(gòu)造一棵二A(B(D),C(E(,G), G 利用二叉樹(shù)的后綴表示來(lái)構(gòu)造一棵二叉 建立一棵二叉樹(shù)的方 利用二叉樹(shù)的后綴表示來(lái)構(gòu)造一棵二叉樹(shù)(習(xí)題(a+b)* ab+(?a+b)* a?b+cd+(-a+b)* da?b+c*利用先序、中序字符串的有關(guān)串的定義、術(shù)語(yǔ)、基本字符串的類部分成員函數(shù)的實(shí)利用前序、中序序列建立字符串（簡(jiǎn)稱串）的定義以及一些*串：是n（n>=0）個(gè)字符的一個(gè)有限序列，開(kāi)頭結(jié) 引號(hào)“”起來(lái)例如B=structure（B為串名*串的長(zhǎng)度：串中所包含的字符個(gè)數(shù)n（不包括分界符‘“’，也不括串的結(jié)束符*空串：長(zhǎng)度為0的串?；蛘哒f(shuō)只包含串結(jié)束符‘\0’的注意\0等于\0空串不等于空*子串：串中任一連續(xù)子序的子pk”不是B的子兩個(gè)串的連接（并置取子求一個(gè)子串在串中第一次出現(xiàn)的位置等Java與C/C++的不Javatr[1]=‖來(lái)進(jìn)行操作。intlength(booleanequals(ObjectobjcharcharAt(intindexStringsubstring(intbeginIndexStringsubstring(intbeginIndex,intendIndexconstintmaxlen=128;classString{String(constString&ob);String(constchar*init);String();~String(){delete[]intLength()const{returnString&operator(intposint //取子intoperator==(constString&ob)returnstrcmp(ch,ob.ch //判別相等否intoperator!=(constString&ob){returnstrcmp(ch,intoperator!()const{returncurlen=String&operator=(constString&ob);//串賦值String&operatorconstString&ob);//并置運(yùn)算char&operator[](inti);intFind(Stringpat)intchar*}部分成員函數(shù)的實(shí)TakingString&String::operator()(intpos,int String*temp=new temp->curLen=0;temp-else{if(pos+len-1>=curLen)len=curLen-temp-for(inti=0,j=pos;i<len;i++,temp-temp-}return}Assigning

String&String::operator=(constString{if{delete[]ch=newstrcpy(ch,ob.ch);}\n‖;return*}4.CreateBinaryTreerecursiveinorder:DBAEGCHFIA CreateBinaryTreerecursivealgorithmvoidCreateBT(Stringpres,ins;BinaryNode<Type>*& intStringprestemp,instempif(pres.length()==0)else{t=newt- while(ins.ch[inpos]!=t->element)CreateBT(prestemp,instemp,t->left);prestemp=pres(inpos+1,pres.length()-instemp=ins(inpos+1,pres.length()-}}CreateBinaryTreerecursivealgorithmBinaryTree(stringpre,stringIn createBT(pre,In,root)}main() HBDFAEKCG‖}CreateBinaryTreerecursivealgorithmBinaryNode<Type>*voidCreateBT(Stringpres,ins{int BinaryNode<Type>*Stringprestemp,instempif(pres.length()==0)returnelse temp=newtemp- while(ins.ch[inpos]!=temp->element)instemp=ins(0,inpos-temp->left=CreateBT(prestemp,prestemp=pres(inpos+1,pres.length()-instemp=ins(inpos+1,pres.length()-temp->right=CreateBT(prestemp,instemp);returntemp;}}CreateBinaryTreerecursivealgorithmpublicBinaryTree(stringpre,string{root=createBT(pre,}main({strings1(―ABHFDECKG‖);BinaryTreet1(s1,s2);preorder(t1);Inorder(t1);}如果已知后序與中序,能否唯一構(gòu)造一棵二叉樹(shù)呢后序 中序 如果已知先序與后序呢ADTandClassPreOutput():outputthedatafieldsinInOutput():outputthedatafieldsinPostOutput():outputthedatafieldsinLevelOutput():outputthedatafieldsinlevelDelete():deleteabinarytree,freeingupitsHeight():returnthetreeSize():returnthenumberofnodesintheADTandClassTheheightofthetreeisdeterminedas:max{hl,hr}+1template<classintBinaryTree<T>::Height(BinaryNode<T>{if(!t)returninthl=Height(t→Left);if(hl>hr)return++hl;elsereturn++hr;}Binary-TreeRepresentationofa樹(shù) 方式：三廣義表表示abfgabfg0122 —右兄弟表示Takeatreeasabinaryaa

c ijclassTreeNode:Tdata;TreeNode*firstchild,*nextsibling;classTree:TreeNode*root,4.8insertnewnodeintemplate<classT>voidTree<T>::Insertchild(T{TreeNode<T>*newnode=newTreeNode<T>(value);if(current->firstchild==NULL)current->firstchild={TreeNode<T>*p=current-while(p->nextsibling!=NULL)p=p->nextsibling;p->nextsibling=newnode;}}4.8 BinaryFGABCDEHIFGABCDEHIJK每棵樹(shù)轉(zhuǎn)為二ABCABCDEFGHIKJ把每棵二叉樹(shù)根用右鏈AABFCGHDIERJBinary AABFCGHDIERJ4.8樹(shù)的遍歷：深度優(yōu)先遍歷，廣度優(yōu)先遍深度優(yōu)先遍先序次序遍歷（先序樹(shù)的根 按先序遍歷根的第一 ，第二 ……等后序次序遍歷（后序 A先根：ABEFCGKLDHIJM 對(duì)應(yīng)的二叉樹(shù)的先序

后根：EFBKLGCHIMJDA對(duì)應(yīng)的二叉樹(shù)的中序一廣度優(yōu)先遍A

4.8D 分 森林的遍深度優(yōu)先遍*先根次序遍F的第一棵樹(shù)的按先根遍歷第一棵樹(shù) 森按先根遍歷其它樹(shù)組成的*中根次序遍按中根遍歷第一棵樹(shù) 森F的第一棵樹(shù)的按中根遍歷其它樹(shù)組成的*后根次序遍按后根遍歷第一棵樹(shù) 森按后根遍歷其它樹(shù)組成的F的第一棵樹(shù)的

二叉樹(shù)的先二叉樹(shù)的中二叉樹(shù)的后K 中根：EFBAIJHKA J后序廣度優(yōu)先遍歷（層次遍歷

ThreadThreadTreeleftThread andrightThreadThreadTreen個(gè)結(jié)點(diǎn)的二叉樹(shù)有2n個(gè)鏈域其中真正有用的是n1個(gè)，其它n1個(gè)都是空前驅(qū)或后繼等運(yùn)算。

ThreadAABCDEFGHJThread

Inorder:^J^JAABCBC^DEF^DEFGHGH機(jī)內(nèi)如一個(gè)結(jié)點(diǎn)增加兩個(gè)標(biāo)記leftThread=rightThread=

leftchild指向 leftchild指向前驅(qū)（某線性序列 rightchild指向 rightchild指向后線索化二叉樹(shù)的 template<classType>class{friendclassintleftThread,rightThread;ThreadNode<Type>*leftchild,*rightchild;Typedata;ThreadNode(constTypeitem):data(item),leftchild(0),rihgtchild(0),leftThread(0),rihgtThread(0){}template<classType>class{//線索二叉樹(shù)的公共ThreadNode<Type>*root;討論線索化二叉樹(shù)的幾個(gè)按中序遍歷中序線遍歷算法（以中序?yàn)槔f歸，非遞歸（需用工作棧）這里前提是中序線索樹(shù)，所以既不要遞歸，也不要棧遍歷算法*找到中序下的第一個(gè)結(jié)點(diǎn)*不斷找后繼如何找后繼p指結(jié)點(diǎn)沒(méi)有 則p=p（p

p有（p->rightThread=則(1)p=ptemplate<classThreadNode<Type>*ThreadInorderIterator<Type>::First({while(current->leftThread==0)current=current->leftchild;returncurrent;}template<classThreadNode<Type>*ThreadInorderIterator<Type>::Next({ThreadNode<Type>*p=current-if(current->rightThread=while(p->leftThread==0)p=p-}template<classvoidThreadInorderIterator<Type>::Inorder({ThreadNode<Type>for(p=Frist();p!=NULL;p=Next())cout<<p->data<<endl;}構(gòu)造中序線索對(duì)已存在的一棵二叉樹(shù)建立中序線索右空域的前驅(qū)、后繼指針。所以除了流動(dòng)指針p外還要加一個(gè)p例如

F F 即p為pre的中序下的后 若pre的右鏈為空，則可填ThreadCreateinorderVoidInthread(threadNode<T>*{stack<threadNode<T>*>sThreadNode<T>*p=T;ThreadNode<T>*pre=NULL;for(;;){1.while{s.push(p);p=p->leftchild;2.if(!s.IsEmpty({1)p=if(pre!={if(pre->rightchild=={pre->rightchild=p;pre->rightthread=1;}if(p->leftchild==NULL){p->leftchild=pre;p->leftthread=1;}pre=p;p=p->rightchild}else}4.8樹(shù)(Huffman增長(zhǎng)樹(shù)的增長(zhǎng)對(duì)原二叉樹(shù)中度為1的結(jié)點(diǎn)，增加一個(gè)空對(duì)原二叉樹(shù)中的樹(shù)葉，增加兩個(gè)空樹(shù)外通路長(zhǎng)度（外路徑）E：根到每個(gè)外結(jié)點(diǎn)（增長(zhǎng)樹(shù)的葉子）內(nèi)通路長(zhǎng)度（內(nèi)路徑）I：根到每個(gè)內(nèi)結(jié)點(diǎn)（非葉子）的路徑長(zhǎng)度的總和（邊結(jié)點(diǎn)的帶權(quán)路徑長(zhǎng)一個(gè)結(jié)點(diǎn)的權(quán)值與結(jié)點(diǎn)的路徑乘積帶權(quán)的外路徑長(zhǎng)葉結(jié)點(diǎn)的帶權(quán)帶權(quán)的內(nèi)路徑長(zhǎng)度：各非葉結(jié)點(diǎn)的帶權(quán)路徑長(zhǎng)樹(shù)給出m個(gè)實(shí)數(shù)W1，W2，…，Wm（m>=2）作為m個(gè)外點(diǎn)的權(quán)構(gòu)造一棵增長(zhǎng)樹(shù)，使得帶權(quán)mWili最小 為wi的外結(jié)點(diǎn)的通路長(zhǎng)例子：外結(jié)點(diǎn)2，3，4，11則可構(gòu)423 42323234算法：從m個(gè)權(quán)值中找出兩個(gè)最小值W1，W2構(gòu)wW=W1+W2表通過(guò)該w的頻度然后對(duì)m-1個(gè)權(quán)值W，W3，W4，…Wm經(jīng)由小到大排序，求解這問(wèn)題例子：23571113171923293137注意:當(dāng)內(nèi)結(jié)點(diǎn)的權(quán)值與外結(jié)點(diǎn)的權(quán)值相等的情況下，內(nèi)結(jié)點(diǎn)應(yīng)外結(jié)點(diǎn)之后。除了保證Wili最小外，還保證maxIj,Ij也有最小例如:789編 樹(shù)在數(shù)據(jù)編碼中一種應(yīng)用。具體的講用于通信的二進(jìn)制編碼中。設(shè)一電文出現(xiàn)的字符為D={M，S，T，A，Q，個(gè)字符出現(xiàn)的頻率為W={10，29，4，8，15，7}何對(duì)上面的諸字符進(jìn)行二進(jìn)制編碼，使得該電文的總長(zhǎng)度最短為了譯碼，任一字符的編碼不應(yīng)是另一字符的編碼的算法：利用Huffman算法，把{10，29，4，8，15，7}作為外部 的邊標(biāo)上0，右 01 001AM1748編碼

已知二進(jìn)制編碼，方法是從根結(jié)點(diǎn)開(kāi)始確定一條到達(dá)葉結(jié)點(diǎn)的路1001100101

111

11010 MS S2009年統(tǒng)考題(單項(xiàng)選擇

Chapter給定二叉樹(shù)如下圖所示設(shè)N代表二叉樹(shù)的根,L代表二叉樹(shù),R代表根結(jié)點(diǎn)的 .若遍歷后的結(jié)點(diǎn)序列為3,1,7,624則其遍歷方A. B. C. D.1 2009年統(tǒng)考題(單項(xiàng)選擇

Chapter,全二叉樹(shù)的結(jié)點(diǎn)個(gè)數(shù)最A(yù). C. D.將森林轉(zhuǎn)換為對(duì)應(yīng)的二叉樹(shù),若在二叉樹(shù)中結(jié)點(diǎn)u是結(jié)點(diǎn)v點(diǎn)的父結(jié)點(diǎn),則在原來(lái)的森林中,u和v可能具有的關(guān)系父子關(guān) 2)兄弟關(guān) 3)u的父結(jié)點(diǎn)與v的父結(jié)點(diǎn)兄弟關(guān)A.只有 B.1)和 1)和 1),2)和Chapter2010 考研3、下列線索二叉樹(shù)中（用虛線表示線索），符合后序線索樹(shù)定義的是 Chapter5、在一棵度為4的樹(shù)T中，若有20個(gè)度為4的結(jié)點(diǎn)，10個(gè)度為3的結(jié)點(diǎn)，1個(gè)為2的結(jié)點(diǎn)，10個(gè)度為1的結(jié)點(diǎn)，則樹(shù)T的葉節(jié)點(diǎn)個(gè)數(shù)是 6、對(duì)n(n大于等于2)個(gè)權(quán)值均不相同的字符構(gòu) 樹(shù)，關(guān)于該樹(shù)的敘中，錯(cuò)誤的是A：該樹(shù)一定是一棵完全二叉 ：樹(shù)中一定沒(méi)有度為1的結(jié) 樹(shù)中兩個(gè)權(quán)值最小的結(jié)點(diǎn)一定是兄弟結(jié)D：樹(shù)中任一非葉結(jié)點(diǎn)的權(quán)值一定不小于下一任一結(jié)點(diǎn)的權(quán)Chapter給出如下各表達(dá)式的二叉1）(a+b)/(c--x-((a+b)>(c-d))||a<f&&(x<y||如果一棵樹(shù)有1個(gè)度為1的結(jié)點(diǎn)，有2個(gè)度為2的結(jié)點(diǎn)，……，個(gè)度為0分別找出滿足以下條件的所有二叉樹(shù)二叉樹(shù)的前序序列與中序序列相二叉樹(shù)的中序序列與后序序列相二叉樹(shù)的前序序列與后序若用二叉鏈表作為二叉樹(shù) 表示，試對(duì)以下問(wèn)題編寫(xiě)遞歸算法統(tǒng)計(jì)二叉樹(shù)中葉結(jié)點(diǎn)的個(gè)數(shù)以二叉樹(shù)為參數(shù)，交換每個(gè)結(jié)點(diǎn)的 和Chapter已知一棵二叉樹(shù)的先序遍列結(jié)果是中序遍列結(jié)果是試畫(huà)出這棵二叉樹(shù)示。設(shè)每個(gè)操作符有一個(gè)或兩個(gè)操給定權(quán)值{15,03,14,02,06,09,16,17},構(gòu)造相應(yīng) 樹(shù)，計(jì)算它的帶權(quán)外路徑長(zhǎng)c1c2c3c4c5c6c7c8這八個(gè)字母的出現(xiàn)頻率{5,25,3,6,10,11,36,4,}為這八個(gè)字母設(shè)計(jì)不等長(zhǎng)的Huffman編碼,給出該電文的總碼數(shù)實(shí)習(xí)題建立一棵二叉樹(shù)，并輸出前序、中序、后序遍歷結(jié)果**General廣義表 結(jié)廣義表的概念

**General*定義為n(n>=0)個(gè)表元素a0a1a2,……an-1組成的有限序列,記作:LS=(a0,a1,a2,……an-1)其中每個(gè)表元素ai可以是原子,也可以是子表原子:某種類型的對(duì)象,在結(jié)構(gòu)上不可分(用小寫(xiě)字母表示).子表:有結(jié)構(gòu)的.(用大寫(xiě)字母表示)*廣義表的長(zhǎng)度:表中元素的個(gè)**General*廣義表的表頭(head),表尾tail=(a1,a2,……an-*廣義表的深度A=(B=(6,

表中所含括號(hào)的最大層C=(?a5,3,?x?))表頭為‘a(chǎn)?,表尾為((5,3,D=(B,C,E=(B,F=(4,F)遞歸的廣義表的性質(zhì)(特點(diǎn)有序有長(zhǎng)度,有深可遞歸,如上面例可共享,如E中B為E,D所共

表示表元素 2626353DBCAEBDC3DBCAEBDCF3562A52644**GeneralGeneralListshead3)first5)next7)pushsetinfosethead

tail4)info6)nodetype8)addonsetnextsettail**GeneralGeneralLists

L=(3,(utyperef01333013333^00^2 2c00^03^03^03^03^2020**General廣義表的#defineHEAD#defineINTGR#defineCH#defineLST3classGenList;{friendclassintGenListNode*union{intintcharcharinfo;GenListNode*hlink;}

**GeneralGenListNode&info(GenListNode*intnodetype(GenListNode*elem){returnelem-voidsetinfo(GenListNode*elem,GenListNode&class{GenListNode*GenListNode*Copy(GenListNode*intdepth(GenListnode*intequal(GenlistNode*s,Genlistnode*voidRemove(GenlistNode*GenList(~GenList(

**GeneralGenListNode&Head();GenListNode*Tail();GenlistNode*First();GenlistNode*Next(GenListNode*voidPush(GenListNode&GenList&Addon(GenList&list,GenListNode&voidsetHead(GenListNode&voidsetNext(GenlistNode*elem1,GenlistNode*voidsetTail(GenList&list);voidCopy(constGenList&l);intdepth();int ist(GenListNode*ls,char*} 2)遞歸函數(shù)的內(nèi)部調(diào) 私有函求廣義表的深

LS==(a0a1a2,……an-1),其中ai(0<=I<=n-1)0

intGenList::depth({return}私有函數(shù)int if