EEPW论坛

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},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7]
  [9,20],
  [3],
]

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

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

思路：层次遍历二叉树，然后再进行逆序操作；

1. class Solution {  

2. public:  

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

4.     {  

5.         vector<vector<int>> res;  

6.   

7.         if (root == NULL)  

8.         {  

9.             return res;  

10.         }  

11.   

12.         queue<TreeNode*> q;  

13.         q.push(root);  

14.   

15.         while (q.size()>0)  

16.         {  

17.             vector<int> level;  

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

19.             {  

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

21.                 q.pop();  

22.   

23.                 level.push_back(tmp->val);  

24.   

25.                 if (tmp->left)  

26.                 {  

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

28.                 }  

29.                 if (tmp->right)  

30.                 {  

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

32.                 }  

33.             }  

34.   

35.             res.push_back(level);  

36.         }  

37.   

38.         reverse(res.begin(), res.end());  

39.   

40.         return res;  

41.     }  

42. };