Populating Next Right Pointers in Each Node II--LeetCode

题目：

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

思路：和上面的一题差不多，只不过需要慢慢的查找，更新的时候也是需要慢慢的查找

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 一直都是上一层的第一个节点 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;                      }          else if(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);     }

ps:这道题的麻烦之处在于查找，需要逐步索引查找。