454647thisLevelList.add(currentNode.val);48if(currentNode.left !=null)49q.add(currentNode.left);50if(currentNode.right !=null)51q.add(currentNode.right);52}5354answerList.add(thisLevelList);55}56returnanswerList;57}58} 还要积累一下java中队列的使用方法: 1.声明和初始化: Queue<TreeNode> q...
import java.util.Queue; import java.util.LinkedList; public class BinaryTreeLevelOrder { public static class TreeNode { int data; TreeNode left; TreeNode right; TreeNode(int data) { this.data=data; } } // prints in level order public static void levelOrderTraversal(TreeNode startNode) {...
* lastNode: 当前层的最后一个节点 * lastInqueue:表示某层最后进入队列的节点 TreeNode curNode = root; TreeNode lastNode = root; TreeNode lastInQueue = root; while(!queue.isEmpty()){ curNode = queue.poll(); listLevel.add(curNode.val); if(curNode.left != null){ queue.offer(curNode...
Binary Tree Level Order Traversal(二叉树的层次遍历) HoneyMoose iSharkFly - 鲨鱼君 来自专栏 · Java 描述给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问)样例给一棵二叉树 {3,9,20,#,#,15,7}:3 / \ 9 20 / \ 15 7返回他的分层遍历结果:[ [3], [9,20], [15,7] ]挑战挑战1...
Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript ...
Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. 栈迭代 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2 ...
题目描述英文版描述Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: Input: root = [3,9,20,null,null,1…
The inOrder traversal is one of the most popular ways to traverse a binary tree data structure in Java. TheinOrdertraversal is one of the three most popular ways to traverse a binary tree data structure, the other two being thepreOrderandpostOrder. During the in-order traversal algorithm, ...
Preorder Traversal: Sample Solution: Java Code: classNode{intkey;Nodeleft,right;publicNode(intitem){// Constructor to create a new Node with the given itemkey=item;left=right=null;}}classBinaryTree{Noderoot;BinaryTree(){// Constructor to create an empty binary treeroot=null;}voidprint_Pre...
Binary Tree Level Order Traversal II是姊妹题,解题思路都是一样的,只是结果要求的顺序是反的,同样有两种方法,也就是经常说到的DFS深度优先遍历和BFS广度优先遍历。 BFS: 广度优先遍历就是一层层地攻略过去,把每一层的所有节点都记录下来再走向下一层。因为每层会有多个节点,不是简单的一个左节点一个右节点的...