199. 二叉树的右视图

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        q = collections.deque([root])
        res = []
        while q:
            size = len(q)
            res.append(q[0].val)
            for _ in range(size):
                node = q.popleft()
                if node.right:
                    q.append(node.right)
                if node.left:
                    q.append(node.left)
        return res

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        if (root == null) return Collections.emptyList();
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        List<Integer> res = new ArrayList<>();
        while (!q.isEmpty()) {
            int size = q.size();
            res.add(q.peek().val);
            while (size-- > 0) {
                TreeNode node = q.poll();
                if (node.right != null) q.offer(node.right);
                if (node.left != null) q.offer(node.left);
            }
        }
        return res;
    }
}