107. Binary Tree Level Order Traversal II

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> list = new LinkedList<>();

        Queue<TreeNode> queue = new LinkedList<>();
        if (root != null) {
            queue.add(root);
        }

        while (!queue.isEmpty()) {
            List<Integer> currentLevelList = new ArrayList<>(queue.size());

            // 当前 queue 中的元素即为当前层的元素
            for (int i = queue.size(); i > 0; i--) {
                TreeNode node = queue.poll();
                currentLevelList.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }

            list.addFirst(currentLevelList); // 添加至链表的头部，以保证最底层的在 List 头部
        }

        return list;
    }
}

Reference

107. Binary Tree Level Order Traversal II