Populating Next Right Pointers in Each Node II [Leetcode]

Problem Description http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

Basic idea is to get every nodes in the current level by iterating nodes of the previous level, then to iterate all nodes in current level to connect them with pointer "next". 

 1 /**

 2  * Definition for binary tree with next pointer.

 3  * struct TreeLinkNode {

 4  *  int val;

 5  *  TreeLinkNode *left, *right, *next;

 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}

 7  * };

 8  */

 9 class Solution {

10 public:

11     void connect(TreeLinkNode *root) {

12         // Start typing your C/C++ solution below

13         // DO NOT write int main() function

14         if(root == NULL)

15             return;

16         if(root->left == NULL && root->right == NULL){

17             root->next = NULL;

18             return;

19         }

20         

21         vector<TreeLinkNode *> nodes_previous_level;

22         nodes_previous_level.push_back(root);

23 

24         while(true) {

25             vector<TreeLinkNode *> nodes_current_level;

26             for(auto item: nodes_previous_level) {

27                 if(item->left != NULL)

28                     nodes_current_level.push_back(item->left);

29                 if(item->right != NULL)

30                     nodes_current_level.push_back(item->right);

31             }

32             

33             if(nodes_current_level.size() == 0)

34                 break;

35             

36             //connect

37             for(int j =0; j< nodes_current_level.size(); j++) {

38                 if(j+1 == nodes_current_level.size())

39                     nodes_current_level[j]->next = NULL;

40                 else

41                     nodes_current_level[j]->next = nodes_current_level[j + 1];

42                 

43             }

44             

45             nodes_previous_level.clear();

46             nodes_previous_level = nodes_current_level;

47         }

48         

49         return;

50     }

51 };