LeetCode Binary Tree Level Order Traversal II

题目：

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) {
        vector<vector<int>> i_vec_vec;
        stack<vector<int>> i_stk_vec;
        if(root == NULL) 
        	return i_vec_vec;
        queue<queue<TreeNode*>> node_que_que;
        queue<TreeNode*> node_que; 
        node_que.push(root);
        node_que_que.push(node_que);
        while(!node_que_que.empty()) {
        	queue<TreeNode*> out_que;
        	queue<TreeNode*> in_que;
        	vector<int> in_vec;
        	out_que = node_que_que.front();
        	while(!out_que.empty()) {
        		TreeNode *tmp = out_que.front();
        		in_vec.push_back(tmp->val);
        		if(tmp->left)
        			in_que.push(tmp->left);
        		if(tmp->right)
        			in_que.push(tmp->right);
        		out_que.pop();   			
        	}
        	node_que_que.pop();
        	i_stk_vec.push(in_vec);
        	if(!in_que.empty())
        		node_que_que.push(in_que); 		
        }
        while(!i_stk_vec.empty()) {
        	vector<int> ivec = i_stk_vec.top();
        	i_stk_vec.pop();
        	i_vec_vec.push_back(ivec);
        	
        }
        return i_vec_vec;
    }
};