[leetcode] Binary Tree Level Order Traversal II

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],

]

与[leetcode] Binary Tree Level Order Traversal的方法相同，利用一个队列来完成层次遍历，最后在把结果反转过来即可。

代码如下：

 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         // Note: The Solution object is instantiated only once and is reused by each test case.

14        

15         vector<vector<int>> result;

16         vector<vector<int>> result2;

17         if(root==NULL) return result;

18         vector<int> temp;

19         queue<TreeNode * > q;

20         q.push(NULL);

21         q.push(root);

22         while(q.size()>1)

23         {

24             if(q.front()==NULL)

25             {

26                 q.pop();

27                 q.push(NULL);

28                 if(temp.size()!=0) result.push_back(temp);

29                 temp.clear();

30             }

31             else

32             {

33                 if(q.front()->left!=NULL)q.push(q.front()->left);

34                 if(q.front()->right!=NULL)q.push(q.front()->right);

35                 temp.push_back(q.front()->val);

36                 q.pop();

37             }

38         }

39         if(temp.size()!=0) result.push_back(temp);

40         for(int i = result.size() - 1; i >= 0; i--) result2.push_back(result.at(i));

41         return result2;

42         

43     }

44 };