[Problem Link]

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       <---
  • Perform a breath first traversal of the tree also called a level order traversal.
  • Before you start processing any level, add the last node from your Q/list to the return list. That’s the node the user would see from the right side.
/**
 * 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> ret = new ArrayList<Integer>();
        LinkedList<TreeNode> q = new LinkedList<TreeNode>();
        if ( root != null )
            q.add(root);
        while ( q.size() > 0 ) {
            int levelSize = q.size();
            ret.add(q.getLast().val);
            for ( int i = 0 ; i < levelSize ; i++ ) {
                TreeNode curr = q.remove(0);
                if ( curr.left != null )
                    q.add(curr.left);
                if ( curr.right != null )
                    q.add(curr.right);
            }
        }
        return ret;
    }
}