( bfs is easy, instead use preorder dfs right first and then left)

 

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       <---



// dfs, preorder, right first and then left
// use a var level to know which level we are on, since
// we only add one element on each level , and 
// the res size is related to the first node on the new level
// for instsance, level 0, list size is 0, and we add cur node at 
// level 0. when we are level one, the list size is 1, 
// we add the cur node to the res, 
// when we meet another new node at level 1, the current list size is 
// 2 , then we know the current node is not the right most node on level 1 
// 
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
      List<Integer> res = new ArrayList<>();
      if(root == null) return res;
      
      dfs(root, res, 0);
      return res;
    }
  
  private void dfs(TreeNode root, List<Integer> res, int level){
    // base case 
    if(root == null) return;
    
    if(level == res.size()){
      res.add(root.val);
    }
    
    dfs(root.right, res, level + 1);
    dfs(root.left, res, level + 1);
        
    }
}



// bfs iterative queue , time o(n), 
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        List<Integer> res = new ArrayList<>();
        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;
    }
}