LeetCode: Populating Next Right Pointers in Each Node II

这题关键是要记录下一层第一个节点

 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     TreeLinkNode *nextright(TreeLinkNode *root) {

12         while (root) {

13             if (root->left) return root->left;

14             if (root->right) return root->right;

15             root = root->next;

16         }

17         return NULL;

18     }

19     void connect(TreeLinkNode *root) {

20         if (!root) return;

21         TreeLinkNode *second, *first;

22         first = root;

23         second = NULL;

24         first->next = NULL;

25         while (first) {

26             if (first->left) {

27                 if (!second) second = first->left;

28                 if (first->right) first->left->next = first->right;

29                 else first->left->next = nextright(first->next);

30             }

31             if (first->right) {

32                 if (!second) second = first->right;

33                 first->right->next = nextright(first->next);

34             }

35             first = first->next;

36             if (!first) {

37                 first = second;

38                 second = NULL;

39             }

40         }

41     }

42 };

 如果可以用extra space，可以用下面这段代码

 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         queue<TreeLinkNode *> S;

13         if (!root) return;

14         S.push(root);

15         S.push(NULL);

16         while (!S.empty()) {

17             TreeLinkNode *front = S.front();

18             S.pop();

19             if (!front) continue;

20             front->next = S.front();

21             if (front->left) S.push(front->left);

22             if (front->right) S.push(front->right);

23             if (!S.front()) S.push(NULL);

24         }

25     }

26 };