LeetCode OJ

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},

return its bottom-up level order traversal as:

[
  [15,7]
  [9,20],
  [3],
]

解题思路：

广度优先遍历，然后对结果进行翻转。

代码：

 1 /**
 2  * Definition for binary tree
 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> > ans;
14         if (root == NULL) return ans;
15         queue<TreeNode*> one;
16         queue<TreeNode*> another;
17         vector<int> cur_ans;
18         TreeNode* cur_node;
19         
20         one.push(root);
21         
22         while (!one.empty() || !another.empty()) {
23             if (!one.empty()) {
24                 while (!one.empty()) {
25                     cur_node = one.front();
26                     one.pop();
27                     cur_ans.push_back(cur_node->val);
28                     if (cur_node->left) another.push(cur_node->left);
29                     if (cur_node->right) another.push(cur_node->right);
30                 }
31                 ans.push_back(cur_ans);
32                 cur_ans.clear();
33             }
34             if (!another.empty()) {
35                 while (!another.empty()) {
36                     cur_node = another.front();
37                     another.pop();
38                     cur_ans.push_back(cur_node->val);
39                     if (cur_node->left) one.push(cur_node->left);
40                     if (cur_node->right) one.push(cur_node->right);
41                 }
42                 ans.push_back(cur_ans);
43                 cur_ans.clear();
44             }
45         }
46         reverse(ans.begin(), ans.end());
47         return ans;
48     }
49 };