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 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) {
vector<vector<int>> result;
if(root == NULL)
return result;
int curNum = 1;
int nextNum = 0;
int level = 0;
//int i = 0;
queue<TreeNode*> qTree;
qTree.push(root);
vector<int> tmp(0);
TreeNode* tNode;
while(!qTree.empty()){
tmp.clear();
for(int i=0;i<curNum;i++){
tNode = qTree.front();
qTree.pop();
tmp.push_back(tNode->val);
if(tNode->left != NULL){
qTree.push(tNode->left);
nextNum++;
}
if(tNode->right != NULL){
qTree.push(tNode->right);
nextNum++;
}
}
curNum = nextNum;
nextNum = 0;
result.push_back(tmp);
}
reverse(result.begin(), result.end());//数组反转
return result;
}
};