Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree{3,9,20,#,#,15,7}, 3/\920/\157 return its level order traversal as: [ [3], [9,20], [15,7] ] 题解:二叉树的层次遍历,用队列...
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) { Queue<TreeNode> queue=ne...
intlv)8:treeNode(node), level(lv){}9};10public:11vector<vector<int>> levelOrder(TreeNode*root) {12queue<Node>nodeQueue;13vector<vector<int>>ret;14if(root ==NULL)15returnret;16nodeQueue.push(Node(root,0));17intdep = -1;18while(!nodeQueue.empty()){19Node node ...
题目: Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree{3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 1. 2. 3. 4. 5. return ...
Binary+Tree+Level+Order+Traversal">https://www.cwiki.us/display/ITCLASSIFICATION/Binary+Tree+Level+Order+Traversal* @seehttps://www.lintcode.com/problem/binary-tree-level-order-traversal* * ** @author YuCheng**/publicclassLintCode0069LevelOrderTest{privatefinalstaticLoggerlogger=LoggerFactory.getLogg...
【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】 原题 Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, ...
Given binary tree [3,9,20,null,null,15,7]. ** 3 / 9 20 / 15 7** return its level order traversal as: [ [3], [9,20], [15,7] ] My Solution (Java) Version 1 Time: 3ms: 队列这个结构就巧妙地把二叉树的立体结构变成线性结构了,也是一种巧妙地遍历方式,先遍历每个小树的根结点,然...
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...
这道题考的就是 BFS,我们可以通过 DFS 实现。只需要在递归过程中将当前 level 传入即可。 publicList<List<Integer>>levelOrder(TreeNoderoot){List<List<Integer>>ans=newArrayList<>();DFS(root,0,ans);returnans;}privatevoidDFS(TreeNoderoot,intlevel,List<List<Integer>>ans){if(root==null){return;}...
In this article, we will discuss 3 different techniques for Level Order Traversal of a binary tree. This technique can be used to find the left and right view of the tree.