Populating Next right pointers in each Node Ii--leetcode

Topic:

Follow up to Problem "populating Next right pointersin each Node".

What if the given tree could is any binary tree? Would your previous solution still work?

Note:

• Constant extra space.

For example,
Given The following binary tree,

         1       /        2    3     /\        4   5    7

After calling your function, the tree is should look like:

         1, NULL       /        2, 3, null     /\        4-> 5, 7, NULL

Idea: And the above question is similar, just need to slowly find, update the time also need to slowly find

void order (bintree* root) {bintree* temp = root;     bintree* next_head = NULL;     if (temp = = NULL) return;                while (temp! = NULL) {cout<<temp->value<<endl;     temp = temp->next;     } temp = root;          while (temp! = null) {if (temp->left! = null) {next_head= temp->left;       Break          } if (Temp->right!=null) {next_head = temp->right;       Break             } temp = temp->next; } order (next_head);}    Pre_head has always been the first node of the previous layer void Helper_third (bintree* pre_head) {bintree* pre = Pre_head;     Bintree* Cur_first=null;    Bintree* Cur_second=null;    Bintree* Next_head=null;         while (pre! = null && (Cur_first = = NULL | | cur_second = = NULL)) {if (pre->left! = null) {           if (Cur_first = = NULL) {cur_first= pre->left;                      Next_head = Cur_first; } elseif (Cur_second ==null) Cur_second = pre->left;           if (pre->right! = null) {if (Cur_first = = null) {Cur_first = pre->right;         Next_head = Cur_first;                    } else if (Cur_second = = NULL) Cur_second = pre->right;      } if (Cur_first! = NULL && Cur_second! = null) break;    Pre = pre->next;    } if (Cur_first = = NULL && Cur_second = = null) return;      while (pre! = NULL) {cur_first->next = Cur_second;       Cur_first = Cur_second;         if (Cur_second = = Pre->left && pre->right! = NULL) {Cur_second = pre->right;      Continue      } pre = pre->next;              while (pre! = null) {if (pre->left! = null) {Cur_second = pre->left;                 Break   } if (pre->right! = NULL) {Cur_second = pre->right;        Break            } pre = pre->next;    }} if (Cur_first! = null) Cur_first->next = NULL; Helper_third (Next_head);}   void Connect_third (bintree* root) {if (root==null) return;   Root->next = NULL;     Helper_third (root); }

Populating Next right pointers in each Node Ii--leetcode