# 107.Binary Tree Level Order Traversal II

难度:Easy

**107.Binary Tree Level Order Traversal II**

> 给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

```
例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
返回其自底向上的层次遍历为：

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

类似于[429题](/leetcode/429.n_ary_tree_level_order_traversal.md)的N叉树的层次遍历。代码如下:

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        deque<TreeNode*> dt;
        vector<vector<int>> res;
        if(!root) return res;
        dt.push_back(root);
        while(!dt.empty())
        {
            vector<int> tmp;
            for(auto i:dt)
                tmp.push_back(i->val);
            int len=dt.size();
            while(len--){
            if(dt[0]->left)
                dt.push_back(dt[0]->left);  
            if(dt[0]->right)
                dt.push_back(dt[0]->right);
            dt.pop_front();
            }
            // cout<<dt.size()<<endl;
            

            res.push_back(tmp);
            
           
        }
        reverse(res.begin(),res.end());
        return res;
        
    }
};
```


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