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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].
层序遍历二叉树，依次输出每层最右节点即可，代码如下：
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> rightSideView(TreeNode root) { List<Integer> re_list = new ArrayList<Integer>(); List<TreeNode> level_list = new ArrayList<TreeNode>();//层序遍历时存储当前层节点的list if(root==null) return re_list; level_list.add(root); re_list.add(root.val); int flag = 1; while(flag==1) { flag = 0; List<TreeNode> tmp_list = new ArrayList<TreeNode>();//用来存储下一层节点的list for(TreeNode node:level_list) { if(node.left!=null) { flag = 1;//说明下层还存在节点 tmp_list.add(node.left); } if(node.right!=null) { flag = 1;//说明下层还存在节点 tmp_list.add(node.right); } } int size = tmp_list.size(); if(tmp_list.size()!=0) { re_list.add(tmp_list.get(size-1).val);//记录当前层最右边节点 level_list.clear(); level_list.addAll(tmp_list);//迭代 } } return re_list; } }
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原文地址：http://www.cnblogs.com/mrpod2g/p/4419130.html
