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> > qh;
        vector<int> arr(0);
        stack<vector<int>> tmpstack;
        queue<TreeNode*> q;
        int cnt=1,i;
        int leaves=0;
        
        TreeNode *p = root;
        if(p==NULL) return qh;
        q.push(p);
        while(!q.empty()){
            arr.clear();
            for(i=0,leaves=0;i<cnt;i++){
                p=q.front();
                arr.push_back(p->val);
                q.pop();
                if(p->left!=NULL) {
                    q.push(p->left);
                    leaves++;
                }
                if(p->right!=NULL) {
                    q.push(p->right);
                    leaves++;
                }
            }
            cnt=leaves;
            tmpstack.push(arr);
        }
        while(!tmpstack.empty()){
            qh.push_back(tmpstack.top());
            tmpstack.pop();
        }
        return qh;
    }
};

Binary Tree Level Order Traversal II