Problem

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 - BFS

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        //save into stack in level order
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                if (i == size-1) res.add(cur.val);
            }
        }
        return res;
    }
}

Solution - DFS

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        dfs(root, 0, res);
        return res;
    }
    private void dfs(TreeNode root, int level, List<Integer> res) {
        if (root == null) return;
        if (res.size() == level) res.add(root.val);
        dfs(root.right, level+1, res);
        dfs(root.left, level+1, res);
    }
}