LeetCode Binary Tree Right Side View

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].

思路分析：这题本质上是考察树的遍历，读完题目要能准备理解题意就是要返回每一个level最右边的结点，所以想到用BFS，借助队列实现，但是每一层从右向左扫描，同时记住index，每一个level index为0的node就是需要保存到结果之中的node。时间复杂度O（N），空间复杂度也是O（N），因为需要额外队列保存结点。

AC Code

/** * 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>();    if(root == null) {        return res;    }        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();    queue.add(root);    while(!queue.isEmpty()){        int numNodeInCurLevel = queue.size();        for(int i = 0; i < numNodeInCurLevel; i++){            TreeNode node = queue.poll();             if(i == 0) res.add(node.val);            if(node.right != null){                queue.add(node.right);            }            if(node.left != null){                queue.add(node.left);            }        }    }    return res;   } }