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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},

    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.

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

1. class Solution {  

2. public:  

3.     vector<vector<int>> zigzagLevelOrder(TreeNode* root)  

4.     {  

5.         vector<vector<int>> res;  

6.         if (root==NULL)  

7.         {  

8.             return res;  

9.         }  

10.   

11.         queue<TreeNode*> q;  

12.         q.push(root);  

13.   

14.         while (q.size())  

15.         {  

16.             vector<int> level;  

17.             for (int i=0, n=q.size(); i<n; i++)  

18.             {  

19.                 TreeNode* tmp = q.front();  

20.                 q.pop();  

21.   

22.                 level.push_back(tmp);  

23.                 if (tmp->left)  

24.                 {  

25.                     q.push(tmp->left);  

26.                 }  

27.                 if (tmp->right)  

28.                 {  

29.                     q.push(tmp->right);  

30.                 }  

31.             }  

32.             res.push_back(level);  

33.         }  

34.   

35.         for (int i=0; i<res.size(); i++)  

36.         {  

37.             if ((i+1)%2==0)  

38.             {  

39.                 reverse(res[i].begin(), res[i].end());  

40.             }  

41.         }  

42.   

43.         return res;  

44.     }  

45. };