}}cout<<endl;preOrderTraverse2(root);//前序遍历 迭代法 输出:1,2,3,4cout<<endl;in...
main() { bitree root; root=CreatTree(); /*调用CreatTree函数构造二叉树*/ printf("preorder traversal:\n"); preorderTraverse(root); /*调用preorderTraverse函数先序遍历二叉树*/ printf("\n"); printf("inorder traversal:\n"); InorderTraverse(root); /*调用inorderTraverse函数中序遍历二叉树*/...
for(i=0;i<=6;i++) printf("%d->",arr[i]); printf("\npreorder traversal of tree\n"); preorder(root); printf("\ninorder traversal of tree\n"); inorder(root); printf("\npostorder traversal of tree\n"); postorder(root); ...
Search code, repositories, users, issues, pull requests... Provide feedback We read every piece of feedback, and take your input very seriously. Include my email address so I can be contacted Cancel Submit feedback Saved searches Use saved searches to filter your results more quickly Ca...
12voidInorderTraversal( BinTree BT )3{4if( BT ) {5InorderTraversal( BT->Left );6/*此处假设对BT结点的访问就是打印数据*/7printf("%d", BT->Data);/*假设数据为整型*/8InorderTraversal( BT->Right );9}10}1112voidPreorderTraversal( BinTree BT )13{14if( BT ) {15printf("%d", BT-...
41、ain()BTNode *root = NULL;printf( 二叉树的遍历 nn);printf( 请按先序递归顺序创建二叉树(结束符 #):n);root = CreatBTree();printf(n 递归先序遍历结果 :n); PreOrder(root);printf(n 非递归中序遍历结果 :n); InOrder(root);printf( 非递归后序遍历结果 :n); PostOrder(root);printf(n ...
preOrder1(root->lchild); preOrder1(root->rchild); } } void inOrder1(BinTree *root) //递归中序遍历 { if(root!=NULL) { inOrder1(root->lchild); cout<<root->data<<" "; inOrder1(root->rchild); } } void postOrder1(BinTree *root) //递归后序遍历 ...
traversals: preorder, inorder, postorder, BFS, DFS Sorting selection insertion heapsort quicksort merge sort Graphs directed undirected adjacency matrix adjacency list traversals: BFS, DFS Even More Knowledge Recursion Dynamic Programming Object-Oriented Programming Design Patterns Combinatorics (n choos...
preOrderRecursion(root1); printf("\n"); //中序遍历树1 printf("该二叉树的中序遍历为:"); midOrderRecursion(root1); printf("\n"); //后序遍历树1 printf("该二叉树的后序遍历为:"); postOrderRecursion(root1); printf("\n");
[i].data); // 统计输出顶点数 count++; // 对此顶点弧表遍历 for (e = GL->adjList[gettop].firstedge; e; e = e->next) { k = e->adjvex; // 将k号顶点邻接点的入度减一 if (!(--GL->adjList[k].in)) { // 若为0则入栈,以便于下次循环输出 stack[++top] = k; } } } //...