LeetCode: Populating Next Right Pointers in Each Node

思路对的，少数次过，基本一次过吧

 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         queue<TreeLinkNode *> S;

15         queue<TreeLinkNode *> T;

16         if (!root) return;

17         S.push(root);

18         while (!S.empty()) {

19             TreeLinkNode *tmp = S.front();

20             S.pop();

21             if (tmp->left || tmp->right) {

22                 T.push(tmp->left);

23                 T.push(tmp->right);

24             }

25             if (!S.empty()) tmp->next = S.front();

26             else tmp->next = NULL;

27             if (S.empty()) {

28                 while (!T.empty()) {

29                     S.push(T.front());

30                     T.pop();

31                 }

32             }

33         }

34     }

35 };

 贴一段后来写的更加好的代码

 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) return;

15         queue<TreeLinkNode*> S, T;

16         S.push(root);

17         while (!S.empty()) {

18             while (!S.empty()) {

19                 TreeLinkNode *tmp = S.front();

20                 S.pop();

21                 if (tmp->left) T.push(tmp->left);

22                 if (tmp->right) T.push(tmp->right);

23                 tmp->next = S.empty()? NULL : S.front();

24             }

25             S.swap(T);

26         }

27     }

28 };