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Given a binary tree, imagine yourself standing on the rightside 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].
/** * 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) { List<Integer> res = new ArrayList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root==null) return res; queue.add(root); while(!queue.isEmpty()) { int levelnum = queue.size(); List<Integer> sublist = new ArrayList<Integer>(); for(int i=0; i<levelnum;i++) { TreeNode node = queue.poll(); if(node.left!=null) queue.add(node.left); if(node.right!=null) queue.add(node.right); sublist.add(node.val); } int size = sublist.size(); res.add(sublist.get(size-1)); } return res; } }
very similar with Binary Tree Level Order Traverse
http://www.cnblogs.com/hygeia/p/4704027.html
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原文地址：http://www.cnblogs.com/hygeia/p/4774753.html
