Binary Tree Level Order Traversal II

来源：Leetcode

原帖：https://oj.leetcode.com/problems/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],
]

代码：

 /**  
  * 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;  
     if (!root) return res;  
     queue<TreeNode*> q;  
     q.push(root);  
     while (!q.empty()) {  
       int size = q.size();  
       vector<int> level;  
       for (int i = 0; i < size; ++i) {  
         TreeNode* node = q.front();   
         q.pop();  
         level.push_back(node->val);  
         if (node->left) {  
           q.push(node->left);  
         }  
         if (node->right) {  
           q.push(node->right);  
         }  
       }  
       res.push_back(level);  
     }  
     // reverse (one more function!)  
     reverse(res.begin(), res.end());  
     return res;  
   }  
 };