LeetCode OJ

题目：

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its zigzag level order traversal as:

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

解题思路：

深度优先遍历

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 
11 class Solution {
12 public:
13     vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
14         vector<vector<int>> ans;
15         if (root == NULL) return ans;
16 
17         vector<TreeNode *> one, another;
18         one.push_back(root);
19 
20         while (!one.empty() || !another.empty()) {
21             if (!one.empty()) {
22                 vector<int> cur_ans;
23                 TreeNode * node = NULL;
24                 for (int i = 0; i < one.size(); i++) {
25                     node = one[i];
26                     cur_ans.push_back(node->val);
27                     if (node->left) another.push_back(node->left);
28                     if (node->right) another.push_back(node->right);
29                 }
30                 ans.push_back(cur_ans);
31                 one.clear();
32             }
33             if (!another.empty()) {
34                 vector<int> cur_ans;
35                 TreeNode * node = NULL;
36                 for (int i = 0; i < another.size(); i++) {
37                     node = another[i];
38                     cur_ans.push_back(node->val);
39                     if (node->left) one.push_back(node->left);
40                     if (node->right) one.push_back(node->right);
41                 }
42                 reverse(cur_ans.begin(), cur_ans.end());
43                 ans.push_back(cur_ans);
44                 another.clear();
45             }
46         }
47         return ans;
48     }
49 };