[Leetcode] Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:

You may only use constant extra space.

 
For example,Given the following binary tree,

         1

       /  \

      2    3

     / \    \

    4   5    7

 
After calling your function, the tree should look like:

         1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \    \

    4-> 5 -> 7 -> NULL

한 문제를 따라가다.

 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          if (root == NULL) {

13             return;

14         }

15         queue<TreeLinkNode* > q;

16         TreeLinkNode *p;

17         int idx = 1, n;

18         q.push(root);

19         while (!q.empty()) {

20             n = idx - 1;

21             idx = 0;

22             p = q.front();

23            if (q.front()->left != NULL) {

24                     q.push(q.front()->left);

25                     idx++;

26                 }

27                 if (q.front()->right != NULL) {

28                     q.push(q.front()->right);

29                     idx++;

30                 }

31             q.pop();

32             for (int i = 0; i < n; ++i) {

33                 p->next = q.front();

34                 if (q.front()->left != NULL) {

35                     q.push(q.front()->left);

36                     idx++;

37                 }

38                 if (q.front()->right != NULL) {

39                     q.push(q.front()->right);

40                     idx++;

41                 }

42                 p = p->next;

43                 q.pop();

44             }

45             p->next = NULL;

46         } 

47     }

48 };

단, 상수의 공간 복잡도만 있다면, 우리는 넥스트 바늘을 이용할 수 있습니다. 왜냐하면 현재 층에 접근할 때, 현재 줄의 넥스트 바늘이 이전 층에 접근할 때 연결되어 있기 때문입니다.

 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         TreeLinkNode *prev, *next;

13         while (root) {

14             prev = nullptr; next = nullptr;

15             for (; root != nullptr; root = root->next) {

16                 if (next == nullptr) next = root->left ? root->left : root->right;

17                 if (root->left) {

18                     if (prev) prev->next = root->left;

19                     prev = root->left;

20                 }

21                 if (root->right) {

22                     if (prev) prev->next = root->right;

23                     prev = root->right;

24                 }

25             }

26             root = next;

27         }

28     }

29 };