[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);
    }
}