标签：
二连水
题目地址：
https://leetcode.com/problems/binary-tree-right-side-view/
题目内容：
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
For example:
Given the following binary tree,
1 <--- / 2 3 <--- \ 5 4 <---
You should return [1, 3, 4].
Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
题目解析：
乍一看，酷炫异常，实际上单纯得让人想哭。
让你找到二叉树每层的最后一个元素。
怎么早？
BFS一层，队列的最后一个节点。
具体代码：
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> rightSideView(TreeNode root) { Deque<TreeNode> bfs = new LinkedList<TreeNode>(); List<Integer> result = new ArrayList<Integer>(); if (root == null) { return result; } bfs.add(root); while (!bfs.isEmpty()) { TreeNode last = bfs.getLast(); result.add(last.val); TreeNode first = bfs.getFirst(); while (first != last) { if (first.left != null) { bfs.add(first.left); } if (first.right != null) { bfs.add(first.right); } bfs.pop(); first = bfs.getFirst(); } // handle last if (last.left != null) { bfs.add(last.left); } if (last.right != null) { bfs.add(last.right); } bfs.pop(); } return result; } }
【原创】leetCodeOj --- Binary Tree Right Side View 解题报告
标签：
原文地址：http://www.cnblogs.com/shadowmydx/p/4770212.html
