[LC] 199. 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.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

   1            <---
 /   
2     3         <---
      
  5     4       <---

Solution 1:
BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root == null) {
            return res;
        }
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (i == 0) {
                    res.add(cur.val);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
            }
        }
        return res;
    }
}

Solution 2:

DFS preOrder

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        helper(root, 0, res);
        return res;
    }
    
    private void helper(TreeNode root, int depth, List<Integer> res) {
        if (root == null) {
            return;
        }
        if (depth == res.size()) {
            res.add(root.val);
        }
        helper(root.right, depth + 1, res);
        helper(root.left, depth + 1, res);
    }
}