Leetcode 199 Binary Tree Right Side View

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.

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

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

Analysis:

method 1:
dfs:
we search the tree with the order of {root, right, left}. We will find that each time we go from the upper level to next level, the first TreeNode we hit is the Node we are looking for.

method 2:
bfs:
When scanning the tree (left -> right) from one level to the next level.
The last node in this level is the node we are looking for.

// dfs
   public List<Integer> rightSideView(TreeNode root) {
	   
	   List<Integer> res = new ArrayList<>();
	   dfsHelper(res, root, 0);
	   return res;
    }
   public void dfsHelper(List<Integer> res, TreeNode root, int level) {
	   if (root == null) return;
	   
	   if (res.size() == level) {
		   res.add(root.val);
	   }
	   
	   dfsHelper(res, root.right, level+1);	   
	   dfsHelper(res, root.left, level+1);	
	   
   }
   
   
   
   // bfs
   public List<Integer> rightSideView1(TreeNode root){
	   
	  List<Integer> res = new ArrayList<>();
	  if (root == null) {
		  return res;
	  }
	  Queue<TreeNode> q = new LinkedList<>();
	  
	  q.offer(root);
	  while(!q.isEmpty()) {
		  int size = q.size();
		  for (int i=0; i<size; i++) {
			  TreeNode cur = q.poll();
			  
			  if (i == size-1) {
				  res.add(cur.val);
			  }
			  
			  if (cur.left != null) {
				  q.offer(cur.left);  
			  }
			  
			  if (cur.right != null) {
				  q.offer(cur.right);
			  }  
		  }	  
	  }
	    
	 return res;  
   }