# 107.Binary Tree Level Order Traversal II(M)

## 1.Description(Medium)

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).

**Example**

Given binary tree `{3,9,20,#,#,15,7}`,

```
    3
   / \
  9  20
    /  \
   15   7
```

return its bottom-up level order traversal as:

```
[
  [15,7],
  [9,20],
  [3]
]
```

## 2.Code

跟69.I思想一样，就是最后再用个arrayList把结果倒过来存一遍。

```
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>();
        Queue<TreeNode> que=new LinkedList<TreeNode>();
        if(root==null){
            return result;
        }

        que.offer(root);
        while(!que.isEmpty()){

            ArrayList<Integer> list=new ArrayList<Integer>();
            int size=que.size();
            for(int i=0;i<size;i++){
                TreeNode current=que.poll();
                list.add(current.val);
                if(current.left!=null){
                    que.offer(current.left);
                }
                if(current.right!=null){
                    que.offer(current.right);
                }
            }
            result.add(list);                      
        }

        ArrayList<ArrayList<Integer>> reverseresult=new ArrayList<ArrayList<Integer>>();
        for(int i=result.size()-1;i>=0;i--){
            reverseresult.add(result.get(i));
        }
        return reverseresult;
    }
```


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