题目：

107. Binary Tree Level Order Traversal II

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,null,null,15,7],

3   / \  9  20    /  \   15   7

 

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]] 题目分析： 这道题本身不难，用一个广度优先遍历先分层，再将每层的结果自左向右输出。方便起见用到了栈和队列。记录这道题是一些语法的使用上，帮助自己回忆。 代码：

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10  #include 
     
      11  #include 
      
       12 class Solution {13 public:14     struct Pair{15         TreeNode* root;16         int level;17         Pair(TreeNode* _root,int _level) : root(_root),level(_level) {}18     };19     vector
       
        
         > levelOrderBottom(TreeNode* root) {20         stack
         
           s;21         queue
          
            q;22 vector
           
            
             > v;23 if(!root) return v;24 q.push(Pair(root,0));25 while(!q.empty())26 {27 if(q.front().root->right)28 q.push(Pair(q.front().root->right,q.front().level+1));29 if(q.front().root->left)30 q.push(Pair(q.front().root->left,q.front().level+1));31 s.push(q.front());32 q.pop();33 }34 int level = s.top().level;35 v=vector
             
              
               >(level+1);36 while(!s.empty())37 {38 v[level-s.top().level].push_back(s.top().root->val);39 s.pop();40 }41 return v;42 }43 };