LeeCode_Populating Next Right Pointers in Each Node II

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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

题目如上所示，这题实在I上进行的追问，第一题提交的太快，导致这道题大意了，原来想用递归的方法求解，但是没有清楚解的扩展方向，闲话不说直接上图：

代码如下：

 class Solution {  public:      void connect(TreeLinkNode *root) {          TreeLinkNode *pLeftMost=root;          while (pLeftMost!=NULL){              TreeLinkNode *ptemp=pLeftMost;              while (ptemp!=NULL){                    connectToNext(ptemp);                    ptemp=ptemp->next;              }              pLeftMost=  findLeftMost(pLeftMost);          }           pLeftMost = root;          while (pLeftMost!=NULL){              TreeLinkNode *tep=pLeftMost;              while (tep!=NULL)              {                  cout<<tep->val<<" ";                  tep=tep->next;              }              cout<<"# ";              pLeftMost=  findLeftMost(pLeftMost);          }       }      TreeLinkNode *findLeftMost(TreeLinkNode *preLeftMost){          while (preLeftMost!=NULL){              if(preLeftMost->left!=NULL)              {                  return preLeftMost->left;              }              if (preLeftMost->right!=NULL)              {                   return preLeftMost->right;              }              preLeftMost=preLeftMost->next;          }          return NULL;      }      void connectToNext(TreeLinkNode *root){          if (root==NULL)          {              return;          }          TreeLinkNode *pSet=root->left;          TreeLinkNode *pSetNext=NULL;          //左儿子节点不空           if (pSet!=NULL){              //如果当前节点的左右儿子都不空，则设左儿子的next节点为右儿子              //同时更改需要设置next的节点为当前节点的右儿子              if(root->right!=NULL){                  pSet->next=root->right;                  pSet=root->right;              }          }          else{              //左儿子节点为空，设需要更改next域的节点为右儿子              pSet=root->right;          }          //当前节点的左右儿子都为空          if (pSet==NULL)          {              /*connectToNext(root->next);*/              return;          }          //find pSet's next           TreeLinkNode *ptemp=root->next;          while (ptemp!=NULL){              if (ptemp->left!=NULL){                  pSetNext=ptemp->left;                  break;              }              if (ptemp->right!=NULL){                  pSetNext=ptemp->right;                  break;              }              ptemp=ptemp->next;          }          pSet->next=pSetNext;        /*  connectToNext(root->next);*/      }  };

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