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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        queue<TreeNode*> q;
        vector<vector<int> >ret;
        if(root==NULL)return ret;
        q.push(root);
        int level=0;
        ret.resize(level+1);
        int count=1;
        int nextCount=0;
        while(q.size()>0){
             if(count==0){
                level++;
                ret.resize(level+1);
                count=nextCount;
                nextCount=0;
            }
            TreeNode* current=q.front();
            ret[level].push_back(current->val);
            q.pop();
            if(current->left!=NULL){
                q.push(current->left);
                nextCount++;
            }
            if(current->right!=NULL){
                q.push(current->right);
                nextCount++;
            }
            count--;
           
        }
        int half=ret.size()/2;
        for(int i=0;i<half;i++){
            vector<int> temp=ret[i];
            ret[i]=ret[ret.size()-i-1];
            ret[ret.size()-i-1]=temp;
        }
        return ret;
    }
};