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],

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：

 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 class Solution {
11 public:
12     vector<vector<int>> levelOrderBottom(TreeNode* root) {
13         vector<vector<int>> lorder;
14         vector<vector<int>> lforder;
15         if(root == NULL)
16         {
17             return lorder;
18         }
19         queue<TreeNode*> q;
20         q.push(root);
21         vector<int> level_tmp2;
22         while(!q.empty())
23         {
24             level_tmp2.clear();
25             queue<TreeNode*> level_tmp1;
26             TreeNode* node = q.front();
27             int size = q.size();
28             
29             for(int i = 0; i < size; i++)
30             {
31                 TreeNode* node = q.front();
32                 q.pop();
33                 if(node->left)
34                 {
35                     level_tmp1.push(node->left);
36                 }
37                 if(node->right)
38                 {
39                     level_tmp1.push(node->right);
40                 }
41                 level_tmp2.push_back(node->val);
42             }
43             while(!level_tmp1.empty())
44             {
45                 q.push(level_tmp1.front());
46                 level_tmp1.pop();
47             }
48             lorder.push_back(level_tmp2);
49         }
50         int n = lorder.size();
51         for(int i = n-1; i >= 0; i--)
52         {
53             lforder.push_back(lorder[i]);
54         }//reverse(lorder.begin(), lorder.end());
55        return lforder; 
56     }
57 };