Binary Tree Level Order Traversal II

/*

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],
]

*/
/**
 * 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
        int impossibleVal = 20000;
        vector<vector<int>> result;
        if(root==NULL) return result;
        TreeNode *divider = new TreeNode(impossibleVal);
        vector<int> line;
        TreeNode *curr;
        queue<TreeNode*> q;
        q.push(root);
        q.push(divider);
        while(!q.empty()){
         curr = q.front();
         q.pop();
         if(curr->val==impossibleVal){
          if(line.size()!=0) result.insert(result.begin(), line);
          line.clear();
          if((!q.empty()) && q.front()->val!=impossibleVal){
            divider = new TreeNode(impossibleVal);
            q.push(divider);
          }
         }
         else {
            if(curr->left) q.push(curr->left);
            if(curr->right) q.push(curr->right);
            line.push_back(curr->val);
         }
        }
        return result;
    }
};