Binary Tree Right Side View

题目：

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].

看到这道题，我首先猜测可能和树的某种遍历方式有关，因为思路就往如何利用遍历的方向上去了。

观察发现，其实从右侧观察，能看到的就是层序遍历的每层最后一个元素。所以没啥好说的~~

public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = q.poll();
                if (i == size - 1) {
                    res.add(cur.val);
                }
                if (cur.left != null) {
                    q.offer(cur.left);
                }
                if (cur.right != null) {
                    q.offer(cur.right);
                }
            }
        }
        return res;
    }