[LeetCode]Binary Tree Zigzag Level Order Traversal

题目解析:(链接)

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题思路：

递归版:

 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>> zigzagLevelOrder(TreeNode* root) {
13         travel(root, 1);
14         bool flag = false;
15         for (auto &i : result) {
16             if (flag) {
17                 reverse(i.begin(), i.end());
18             }
19             flag = !flag;
20         }
21         
22         return result;
23     }
24     
25     void travel(TreeNode *root, int level) {
26         if (!root) return;
27         
28         if (level > result.size()) {
29             result.push_back(vector<int>());
30         }
31         
32         result[level - 1].push_back(root->val);
33         travel(root->left, level + 1);
34         travel(root->right, level + 1);
35     }
36 private:
37     vector<vector<int>> result;
38 };

迭代版:

 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>> zigzagLevelOrder(TreeNode* root) {
13         vector<vector<int>> result;
14         if (!root) return result;
15         
16         queue<TreeNode *> current, next;
17         vector<int> level;
18         current.push(root);
19         while (!current.empty()) {
20             while(!current.empty()) {
21                 TreeNode *tmp = current.front();
22                 current.pop();
23                 level.push_back(tmp->val);
24                 
25                 if (tmp->left != nullptr) next.push(tmp->left);
26                 if (tmp->right != nullptr) next.push(tmp->right);
27             }
28             
29             result.push_back(level);
30             level.clear();
31             swap(current, next);
32         }
33         
34         bool flag = false;
35         for (auto &i : result) {
36             if (flag) {
37                 reverse(i.begin(), i.end());
38             } 
39             flag = !flag;
40         } 
41         
42         return result;
43     }
44 };