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.

//Keywords: queue, terminal pointer(mark the end of each level)
//Functions: queue.push(), queue.pop(), queue.front()
//
/**
 * 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>> re;
        vector<int> thisRe;
        queue<TreeNode*> q;
        if(root == NULL){
            return re;
        }
        q.push(root);
        TreeNode * levelEndNode = root;
        while(!q.empty()){
            TreeNode * p = q.front();
            q.pop();
            if(p->left){
                q.push(p->left);
            }
            if(p->right){
                q.push(p->right);
            }
            thisRe.push_back(p->val);
            if(p == levelEndNode){
                //new level
                levelEndNode = q.back();
                re.insert(re.begin(), thisRe);
                thisRe.clear();
            }
        }
        return re;
    }
};