Binary Tree Level Order Traversal II-LeetCode

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) {
       	queue <TreeNode *> m_t;
	    vector <int> lresult;
	    vector < vector <int> > result;
	    TreeNode *temp;
    	int lnum = 0;
    	if (root == NULL)
    	{
    		return result;
    	}
    	m_t.push(root);
    	while(!m_t.empty())
    	{
    		lnum = m_t.size();
	    	for (unsigned int i = 0; i < lnum; i++)
	    	{
	    		temp = m_t.front();
	    		lresult.push_back(temp->val);
	    		if (temp->left != NULL)
	    			m_t.push(temp->left);
    			if (temp->right != NULL)
    				m_t.push(temp->right);
    			m_t.pop();
    		}
    		result.push_back(lresult);
    		lresult.clear();
    	}//此时，输出结果为从根开始一直到叶子节点
    	vector < vector <int> > result1;//
    	for(int i = result.size() - 1; i >= 0; i--)
    	{
    		result1.push_back(result[i]); 
    	}
    	result.clear();
	return result1;//此时输出的结果为从叶子节点开始，一直到根
    }
};