Here is soln for Binary Tree Right Side View where trying to solve below problem

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

Input: root = [1,null,3] Output: [1,3]

Code with comments provided below

            var rightSideView = function(root) {
              const levels = [];
            //DFS solutions often allow us to find a concise, recursive solution, and while they're not always the first thought when it comes to tree traversal problems where the level is important, in this case we don't need the level as a whole, we just need one end of each level. 
              dfs(root, levels)
              const res = [];
              for(let l of levels){
                res.push(l.pop())
              }
              return res
            };

            function dfs(root, levels, level = 0){
             // Base Case
              if(!root) return;
              if(!levels[level]){
                levels[level] = [];
              }
              levels[level].push(root.val);
            // Recur for left subtree then right subtree 

              dfs(root.left, levels, level + 1)
              dfs(root.right, levels, level + 1)
            }

Questions are :-

1. The line if(!root) return; What does it return exactly
2. Does the comments provided in code are right?

Your response is much appreciated

Regards

Carolyn