107. Binary Tree Level Order Traversal II(二叉树广度遍历，自底向上输出)

problems：
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]
]
tip:
广度遍历使用队列实现，vector的反转实现自底向上输出。
solution:
迭代法：

 class Solution {
    public:
        vector<vector<int>> levelOrderBottom(TreeNode* root) {
            vector<vector<int>> res;
            queue<TreeNode *> q{{root}};
            while(!q.empty())
            {
                vector<int> vec;
                for(int i=q.size();i>0;i--)
                {
                    TreeNode *t=q.front();
                    q.pop();
                    vec.push_back(t->val);
                    if(t->left) q.push(t->left);
                    if(t->right) q.push(t->right);
                }
                res.push_back(vec);
            }
            reverse(res.begin(),res.end());//reverse()实现vector反转
            return res;
        }
    };

**递归法：**引入变量level来标记当前深度，即层数，

class Solution {
public:
   
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
       vector<vector<int>> res;
       dfs(root,0,res);
       reverse(res.begin(),res.end());
       return res;
    }
    void dfs(TreeNode *root,int level,vector<vector<int>>& res)//这里注意是vector的引用
    {
        if(!root) return;
        if(res.size() == level) res.push_back({});//当level等于向量大小时申请新的一层
        res[level].push_back(root->val);
        if(root->left) dfs(root->left,level+1,res);
        if(root->left) dfs(root->right,level+1,res);
    } 
};