# 107. Binary Tree Level Order Traversal II

### BFS

```java
/**
 * 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<List<Integer>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> result = new LinkedList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int qsize = queue.size();
            List<Integer> list = new ArrayList<>();
            for (int q = 0; q < qsize; q++) {
                TreeNode cur = queue.poll();
                if (cur == null) {
                    continue;
                }
                list.add(cur.val);
                queue.offer(cur.left);
                queue.offer(cur.right);
            }
            if (!list.isEmpty()) {
                result.offerFirst(list);
            }
        }
        return result;
    }
}
```

### DFS

```java
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> result = new LinkedList<>();
        dfs(root, 0, result);
        return result;
    }
    private void dfs(TreeNode node, int level, LinkedList<List<Integer>> result) {
        if (node == null) {
            return;
        }
        if (level >= result.size()) {
            result.offerFirst(new LinkedList<>()); // add new list in the front side
        }
        result.get(result.size() - level - 1).add(node.val); // almost like, always add in head one
        dfs(node.left, level+1, result);
        dfs(node.right, level+1, result);
    }
}
```
