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<int> tmp;
        queue<TreeNode *> Qpre,Qcurr;
        if(root==NULL)return res;
        else
        {
            Qpre.push(root);
            while(1)
            {
                while(!Qpre.empty())
                {
                  TreeNode *tem=Qpre.front();
                  Qpre.pop();
                  tmp.push_back(tem->val);
                  if(tem->left)Qcurr.push(tem->left);
                  if(tem->right)Qcurr.push(tem->right);
                }
                res.push_back(tmp);
                tmp.clear();
                if(Qcurr.empty())break;
                else Qcurr.swap(Qpre);
            }
            reverse(res.begin(),res.end());
            return res;
        }        
    }
};