答案
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

题目

class Solution {
    public List<Integer> largestValues(TreeNode root) {
        if(root == null) return new ArrayList<Integer>(); ;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        List<Integer> list = new ArrayList<>();
        q.offer(root);
        while(q.size() != 0) {
            int curr_size = q.size();
            int max = Integer.MIN_VALUE;
            for(int i = 0; i < curr_size; i++) {
                TreeNode t = q.poll();
                if(t.val > max)
                    max = t.val;
                if(t.left != null) q.offer(t.left);
                if(t.right != null) q.offer(t.right);
            }
            list.add(max);
        }
        return list;
    }
}