數(shù)據(jù)結(jié)構(gòu)實(shí)訓(xùn)報(bào)告

1、1課程設(shè)計(jì)目標(biāo)與任課程設(shè)計(jì)目課程設(shè)計(jì)任課程設(shè)計(jì)要分析與設(shè)2題目分結(jié)構(gòu)設(shè)算法描程序流程測(cè)試1課程設(shè)計(jì)目標(biāo)與任課程設(shè)計(jì)目課程設(shè)計(jì)任課程設(shè)計(jì)要分析與設(shè)2題目分結(jié)構(gòu)設(shè)算法描程序流程測(cè)試程序說(shuō)3程4測(cè)4.1 測(cè)試數(shù)分5總參考文11.21.311.21.322.1結(jié)2- #22.1結(jié)2- #defineMAX_TREE_SIZEtypedef structPTnode; typedefPTNode 孩子結(jié)點(diǎn)：typedefstructstruct 雙親結(jié)點(diǎn)：typedef Elem typedefPTnode; typedefPTNode 孩子結(jié)點(diǎn)：typedefstructstruct 雙親結(jié)點(diǎn)：typ

2、edef Elem typedef CTBoxn, 個(gè)結(jié)點(diǎn)的兩個(gè)指針?lè)謩e指向其第一個(gè)孩子結(jié)點(diǎn)和下一個(gè)兄弟2.32-62-72-82.32-62-72-8。時(shí)，是分別遍歷以T-和T-rchild ( T )visite( ( BiTree (T-lchild)T-rchild)一棵樹（或樹形）是一個(gè)有限非空的結(jié)點(diǎn)集合T，（2）其余結(jié)點(diǎn)被分成 個(gè)不。時(shí)，是分別遍歷以T-和T-rchild ( T )visite( ( BiTree (T-lchild)T-rchild)一棵樹（或樹形）是一個(gè)有限非空的結(jié)點(diǎn)集合T，（2）其余結(jié)點(diǎn)被分成 個(gè)不相交的集合 T1，T2，,Tm,且又都是樹樹 T1，T2，,

3、Tm,成為每一棵樹的根都root(T)有一條邊相連。voidCreateTree(TreeNode*TreeNode=TreeNode*b TreeNodeTreeNoded-child0=TreeNode*c= root=newTreeNode(A,root-child0d; =root-root-child22.4叉樹的結(jié)構(gòu)體，開辟內(nèi)存空間，借助隊(duì)列實(shí)現(xiàn)轉(zhuǎn)化，其程序流程圖如下2-92.4叉樹的結(jié)構(gòu)體，開辟內(nèi)存空間，借助隊(duì)列實(shí)現(xiàn)轉(zhuǎn)化，其程序流程圖如下2-92.5TreeNode *binaryRoot = rreturn void CreateTree(TreeNode* TreeNode

4、=2.5TreeNode *binaryRoot = rreturn void CreateTree(TreeNode* TreeNode =TreeNode*g TreeNode d-child0= g; TreeNode*c=newroot = new TreeNode(A, root-=3#includecharstructTreeNode*child3;/孩子的數(shù)組 TreeNode(char element = childNumbers=3#includecharstructTreeNode*child3;/孩子的數(shù)組 TreeNode(char element = childNum

5、bers=for i=0; i3; childi = TreeNode& operator = (const TreeNode& if(this=&other) return *this;element = other.element; childNumbers=other.childNumbers;for i=0; ichild0= g; TreeNode*c=newroot = new TreeNode(A, root-child0=root-BTreeNode*TreeToBinaryTree(TreeNode if (treeRoot = return BTreeNode*binary

6、RootnewBTreeNode; binaryRoot-element = treeRoot-element;binaryRoot-leftTreeToBinaryTree(treeRoot-child0); BTreeNode *brotherChild = binaryRoot-left; /兄弟for i= 1; ichildNumbers; brotherChild-right=TreeToBinaryTree(treeRoot-childi); brotherChild = brotherChild-right;return void MiddleOrd(BTreeNode BTreeNode*temp=root; if(temp = NULL)rrTreeNode *treeRoot; return0;*binaryRoot=rrrTreeNode *treeRoot; return0;*binaryRoot=r44.1AABCDBCEGDEFF4-1始G4-2叉F、B、C、G、D、A，故此次程序運(yùn)行理論的結(jié)果為E、F、B、C、G、D、A44.1AABCDBCEGDEFF4-1始