item = parts[0]; TreeNode root = new TreeNode(Integer.parseInt(item)); Queue nodeQueue...= new LinkedList(); nodeQueue.add(root); int index = 1; while(!...nodeQueue.isEmpty()) { TreeNode node = nodeQueue.remove(); if (index ==...leftNumber = Integer.parseInt(item); node.left = new TreeNode(leftNumber); nodeQueue.add...rightNumber = Integer.parseInt(item); node.right = new TreeNode(rightNumber); nodeQueue.add
root) return result; queue nodeQueue; TreeNode *node = root; nodeQueue.push...nodeQueue.empty()) { queue::size_type size = nodeQueue.size(); //每次依次取出这一层的节点...if (node->right) nodeQueue.push(node->right); if (node->left) nodeQueue.push(node->left);...(nodeQueue.Count !...= null) nodeQueue.Enqueue(node.right); if (node.left !
root) { return ans; } queue nodeQueue; nodeQueue.push...nodeQueue.empty()) { deque levelList; int size = nodeQueue.size(); ...for (int i = 0; i < size; ++i) { auto node = nodeQueue.front(); nodeQueue.pop...levelList.push_front(node->val); } if (node->left) { nodeQueue.push...(node->left); } if (node->right) { nodeQueue.push
nodeQueue.push(root); TreeLinkNode *treeNode = nullptr; //二叉树层次遍历 while...nodeQueue.empty()) { treeNode = nodeQueue.front(); nodeQueue.pop();...if (treeNode->right) nodeQueue.push(treeNode->right); } vector::size_type...nodeQueue.isEmpty()) { treeNode = nodeQueue.poll(); nodeList.add(treeNode);...= null) nodeQueue.offer(treeNode.left); if (treeNode.right !
nodequeue.push(root); while(!...nodequeue.empty()) //建立节点队列,打印父节点,入队左右子节点,出队父节点 { node* p = nodeQueue.front();...= NULL) nodeQueue.push(p->right); nodeQueue.pop(); } } 每个节点最多会被访问 2 次, 所以遍历操作的时间复杂度...nodequeue.empty()) //建立节点队列,打印父节点,入队左右子节点,出队父节点 { node* p = nodequeue.front(...= NULL) nodequeue.push(p->right); nodequeue.pop(); }
; nodeQueue.push(nodes[0]); TreeNode *node; int index = 1; while (index < size) { node = nodeQueue.front...(); nodeQueue.pop(); nodeQueue.push(nodes[index++]); node->left = nodeQueue.back(); nodeQueue.push...(nodes[index++]); node->right = nodeQueue.back(); } return nodes[0]; } ---- 下面是一个测试代码,我们可以看看结果: 头文件声明...; nodeQueue.push(nodes[0]); TreeNode *node; int index = 1; while (index < size) { node = nodeQueue.front...(); nodeQueue.pop(); nodeQueue.push(nodes[index++]); node->left = nodeQueue.back(); nodeQueue.push
); LinkedList level_list = new LinkedList(); LinkedList nodeQueue...= new LinkedList(); nodeQueue.add(root); nodeQueue.add(null); while (!...nodeQueue.isEmpty()) { TreeNode treeNode = nodeQueue.poll(); if (treeNode !...= null) nodeQueue.add(treeNode.left); if (treeNode.right !...() > 0) nodeQueue.addLast(null); isleftToRight = !
= new LinkedList(); nodeQueue.addLast(root); nodeQueue.addLast(null); LinkedList...() > 0) { TreeNode currNode = nodeQueue.pollFirst(); if (currNode !...= null) { nodeQueue.addLast(currNode.left); } if...= null) { nodeQueue.addLast(currNode.right); } } else...() > 0) { nodeQueue.addLast(null); } isOrderLeft
this.start=start; for (int i=0;i<10000;i++){ dist[i]=INF; } Queue nodeQueue...=new PriorityQueue(); nodeQueue.add(new Node(start,0)); dist[start]=0; path...nodeQueue.isEmpty()) { Node node = nodeQueue.poll(); int u = node.to;...&dist[i]>arr[u][i]+node.cost){ dist[i]=arr[u][i]+node.cost; nodeQueue.add
const { return a->weight > b->weight;//小顶堆 } }; priority_queue,compareNode> nodeQueue...const { return a->weight > b->weight; //小顶堆 } }; priority_queue,compareNode> nodeQueue...tmp=NULL; for(int i=0;i<len;i++){ initNewNode(tmp); tmp->weight=input[i]; nodeQueue.push...()>1){ leftPtr = nodeQueue.top(); nodeQueue.pop(); rightPtr = nodeQueue.top(); nodeQueue.pop...(); getNewFatherNode(father,leftPtr,rightPtr); currRoot = father; nodeQueue.push
if(root == null){ return 0; } int sum = 0; Queue nodeQueue...= new LinkedList(); Queue numQueue = new LinkedList(); nodeQueue.offer...nodeQueue.isEmpty()){ TreeNode node = nodeQueue.poll(); int num = numQueue.poll...= null){ nodeQueue.offer(left); numQueue.offer(num * 10 + left.val...= null){ nodeQueue.offer(right); numQueue.offer(num * +right.val
= new LinkedList(); nodeQueue.offer(root); boolean isOrderLeft = true;...nodeQueue.isEmpty()) { Deque levelList = new LinkedList();...int size = nodeQueue.size(); for (int i = 0; i < size; ++i) { TreeNode curNode...= nodeQueue.poll(); if (isOrderLeft) { levelList.offerLast(curNode.val...= null) { nodeQueue.offer(curNode.right); } }
; //使用C++的STL标准模板库 nodeQueue.push(root); Node *node; while(!...nodeQueue.empty()){ node = nodeQueue.front(); nodeQueue.pop(); printf(format,...node->data); if(node->lchild){ nodeQueue.push(node->lchild); //先将左子树入队...: //广度优先遍历 void breadthFirstSearch(Tree root){ queue nodeQueue; //使用C++的STL标准模板库 nodeQueue.push...nodeQueue.empty()){ node = nodeQueue.front(); nodeQueue.pop(); printf(format,
*/ public void createHuffman(int[] weights) { // 构建小根堆,弹出是最小的元素 Queue nodeQueue...for (int i = 0; i < weights.length; i++) { nodes[i] = new Node(weights[i]); nodeQueue.add...(nodes[i]); } // 节点队列只剩一个节点结束,即根节点构建完成 while (nodeQueue.size() > 1) {...// 弹出权值最小的两个结点 Node left = nodeQueue.poll(); Node right = nodeQueue.poll();...(parent); } // 哈夫曼树根,遍历要用,优先级队列GC root = nodeQueue.poll(); // 实现编码
= new LinkedList(); nodeQueue.offer(root); boolean isOrderLeft = true;...nodeQueue.isEmpty()) { // 双端队列是一个可以在队列任意一端插入元素的队列。...Deque levelList = new LinkedList(); int size = nodeQueue.size();...for (int i = 0; i < size; ++i) { TreeNode curNode = nodeQueue.poll();...= null) { nodeQueue.offer(curNode.right); } }
(); if (root == null) { return paths; } Queue nodeQueue...= new LinkedList(); Queue pathQueue = new LinkedList(); nodeQueue.offer...nodeQueue.isEmpty()) { TreeNode node = nodeQueue.poll(); String path = pathQueue.poll...= null) { nodeQueue.offer(node.left); pathQueue.offer(new StringBuffer...= null) { nodeQueue.offer(node.right); pathQueue.offer(new StringBuffer
层序遍历: 1 // 利用队列先进先出的特性, 层层输出 2 public void LevelTraversal() 3 { 4 Queue nodeQueue...= null) { nodeQueue.Enqueue(root); } 6 7 while (nodeQueue.Count() !...= 0) 8 { 9 Node currentNode = nodeQueue.Dequeue(); 10 Console.Write...= null) { nodeQueue.Enqueue(currentNode.left); } 13 if (currentNode.right !...= null) { nodeQueue.Enqueue(currentNode.right); } 14 } 15 } 5.
; //构建哈夫曼树 public void createHuffmanTree(int[] weights) { //优先队列,用于辅助构建哈夫曼树 Queue nodeQueue...(nodes[i]); } //主循环,当结点队列只剩一个结点时结束 while (nodeQueue.size() > 1) { //从结点队列选择权值最小的两个结点...Node left = nodeQueue.poll(); Node right = nodeQueue.poll(); //创建新结点作为两结点的父节点 Node parent...= new Node(left.weight + right.weight, left, right); nodeQueue.add(parent); } root...= nodeQueue.poll(); } //输入字符下表,输出对应的哈夫曼编码 public String convertHuffmanCode(int index) {
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