Problem

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

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

思路一(BFS)

1. 如 I,
2. 使用linkedList, 从上往下把level加到list的前面

    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result = new LinkedList<>();
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new LinkedList<>();
            for (int i = 0; i < size; i++) {
                TreeNode curt = queue.poll();
                list.add(curt.val);
                if (curt.left != null) {
                    queue.offer(curt.left);
                }
                if (curt.right != null) {
                    queue.offer(curt.right);
                }
            }
            result.add(0, list);
        }
        return result;
    }

思路二(DFS)

1. 记录层数, 从1开始
2. 如果当前层比list的size要大的话, list增加一个. 因为到最后, list的大小和树的层数恰好相同
3. 把当前节点的值加到它所在层数对应于list里面的位置

        public List<List<Integer>> levelOrderBottom(TreeNode root) {
            List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
            levelMaker(wrapList, root, 1);
            return wrapList;
        }

        public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
            if(root == null) return;
            if(level > list.size()) {
                list.add(0, new LinkedList<Integer>());
            }
            list.get(list.size()-level).add(root.val);
            levelMaker(list, root.left, level+1);
            levelMaker(list, root.right, level+1);
        }