LeetCode--Binary Tree Level Order Traversal II

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/** * Definition for binary tree * 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)     {        vector<vector<int>> res;        vector<vector<int>> temp;        if(root==NULL)            return res;        queue<TreeNode*> use;        use.push(root);        int save = 1;        while(save!=0)        {            int count = save;            save = 0;            vector<int> leve;            while(count>0)            {                TreeNode* tree = use.front();                use.pop();                count--;                leve.push_back(tree->val);                if(tree->left != NULL)                {                    save++;                    use.push(tree->left);                }                if(tree->right!=NULL)                {                    save++;                    use.push(tree->right);                }            }            temp.push_back(leve);        }        for(int i=temp.size()-1; i>=0; i--)            res.push_back(temp[i]);        return res;    }};