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].

给定一颗二叉树，输出它的right view，所谓right view就是从右边观察这棵树所能看到的节点。很容易想到的是用广搜，记录每一层最后一个节点。另外我们还可以用深度优先搜索，从树的右子树开始遍历，不过要维护一个变量layer，来判定当前节点是否应该能看到，因为如果右子树为空后，我们要观察左子树，因为左子树的深度可能比右子树大，我们在观察左子树时，通过layer与已经记录的右子树的节点比较，如果layer大于已经记录的节点数，那么这个节点就可以被观察到，如果小于或等于就不能被观察到。下面给出两种方法的代码：
1，广度优先搜索

/**
 * 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> list = new ArrayList<Integer>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if(root == null) return list;
        queue.offer(root);
        int count = 1;
        int helper = 0;
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            count --;
            if(node.left != null) {
                queue.offer(node.left);
                helper ++;
            }
            if(node.right != null) {
                queue.offer(node.right);
                helper ++;
            }
            if(count == 0) {
                list.add(node.val);  
                count = helper;
                helper = 0;
            }
        }
        return list;
    }
}

2，深度优先搜索

/**
 * 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> list = new ArrayList<Integer>();
        if(root == null) return list;
        getView(root, list, 1);
        return list;
    }
    public void getView(TreeNode root, List<Integer> list, int layer) {
        if(root == null) return;
        if(layer > list.size()) list.add(root.val);
        getView(root.right, list, layer + 1);
        getView(root.left, list, layer + 1);
    }
}