LeeCode_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

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);*/
}
};