Leetcode: Binary Tree Right Side View

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.

For example:
Given the following binary tree,
1            <---
/   \
2     3         <---
\     \
5     4       <---
You should return [1, 3, 4].

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
ArrayList<Integer> res = new ArrayList<Integer>();
if (root == null) return res;
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int PNum = 1;
int CNum = 0;
while (!queue.isEmpty()) {
TreeNode cur = queue.poll();
PNum--;
if (cur.left != null) {
queue.offer(cur.left);
CNum++;
}
if (cur.right != null) {
queue.offer(cur.right);
CNum++;
}
if (PNum == 0) {
res.add(cur.val);
PNum = CNum;
CNum = 0;
}
}
return res;
}
}

public class Solution {
public List<Integer> rightSideView(TreeNode root) {
// reverse level traversal
List<Integer> result = new ArrayList();
Queue<TreeNode> queue = new LinkedList();
if (root == null) return result;

queue.offer(root);
while (queue.size() != 0) {
int size = queue.size();
for (int i=0; i<size; i++) {
TreeNode cur = queue.poll();
if (i == 0) result.add(cur.val);
if (cur.right != null) queue.offer(cur.right);
if (cur.left != null) queue.offer(cur.left);
}

}
return result;
}
}