private void traverseQueue(BinTreeNode<Type> root, NodeHandler<Type> handle) { LinkQueue<BinTreeNode<Type>> queue = new LinkQueue<BinTreeNode<Type>>(); BinTreeNode<Type> node; queue.enQueue(root); while ((node = queue.deQueue()) != null) { if (node.left != null) queue.enQueue(n...
publicfinalbooleanhasQueuedPredecessors() {//The correctness of this depends on head being initialized//before tail and on head.next being accurate if the current//thread is first in queue.Node t = tail;//Read fields in reverse initialization orderNode h =head; Node s;returnh != t &&((s...
HashSet的一个子类,一个链表。 9、SortedSet API---ASetthat further provides atotal orderingon its elements. The elements are ordered using theirnatural ordering, or by aComparatortypically provided at sorted set creation time. The set's iterator will traverse the set in ascending element order. ...
The Iterator provided in method #iterator() and the Spliterator provided in method #spliterator() are not guaranteed to traverse the elements of the priority queue in any particular order. If you need ordered traversal, consider using Arrays.sort(pq.toArray()). Note that this implementation is...
queue.arrays[queue.rear] = value; // rear节点移动到新的无效元素位置上 queue.rear = (queue.rear + 1) % queue.arrays.length; } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 1.4.4 遍历 只要front节点不指向rear节点,就可以一直输出 public static void traverseQueue(Queue queue) { ...
The Iterator provided in method #iterator() and the Spliterator provided in method #spliterator() are not guaranteed to traverse the elements of the PriorityBlockingQueue in any particular order. If you need ordered traversal, consider using Arrays.sort(pq.toArray()...
public final boolean hasQueuedPredecessors() {// The correctness of this depends on head being initialized// before tail and on head.next being accurate if the current// thread is first in queue.Node t = tail; // Read fields in reverse initialization orderNode h = head;Node s;return h ...
正确答案: A C 你的答案: B D(错误)ArrayBlockingQueue：基于数组，在创建ArrayBlockingQueue对象时...
traverse(root->right); //后续遍历:先后续访问左子树,再访问右子树,再访问根节点 } 1. 2. 3. 4. 5. 6. 7. 8. 9. 一、二叉树的前序遍历:迭代和递归 class Solution { public: //vector<int> result;//递归的话定义在这里 vector<int> preorderTraversal(TreeNode* root) { ...
public void levelTraverse(TreeNode root) { if (root == null) { return; } LinkedList<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); System.out.print(node.val+" "); if (node.left != null) { queue.offer(node...