leetcode 107

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]
]
在102题基础上对输出结果进行反转操作，重新定义了一个vector，将102的结果倒序存入vector中。

代码如下：

 /**
  * Definition for a binary tree node.
  * 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>> lorder;
         vector<vector<int>> lforder;
         if(root == NULL)
         {
             return lorder;
         }
         queue<TreeNode*> q;
         q.push(root);
         vector<int> level_tmp2;
         while(!q.empty())
         {
             level_tmp2.clear();
             queue<TreeNode*> level_tmp1;
             TreeNode* node = q.front();
             int size = q.size();

             for(int i = ; i < size; i++)
             {
                 TreeNode* node = q.front();
                 q.pop();
                 if(node->left)
                 {
                     level_tmp1.push(node->left);
                 }
                 if(node->right)
                 {
                     level_tmp1.push(node->right);
                 }
                 level_tmp2.push_back(node->val);
             }
             while(!level_tmp1.empty())
             {
                 q.push(level_tmp1.front());
                 level_tmp1.pop();
             }
             lorder.push_back(level_tmp2);
         }
         int n = lorder.size();
         for(int i = n-; i >= ; i--)
         {
             lforder.push_back(lorder[i]);
         }//reverse(lorder.begin(), lorder.end());
        return lforder;
     }
 };