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].

这道题就是BT的Level Order Traversal，每次要换一层的时候，记录当前节点

/**
 * 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;
    }
}

 注意，每次都是先添加右边child，所以是i==0的时候add

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;
    }
}