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

1. Iterative: 非常巧妙地用了prev和next！

class Solution {
public:
    void connect(TreeLinkNode *root) {
       
        while(root)
        {
            TreeLinkNode *prev = NULL;
            TreeLinkNode *next = NULL;
           
            for ( ; root;root=root->next){
                if (next ==NULL)
                   next = root->left ? root->left : root->right;
                  
                if (root->left){
                    if (prev)
                        prev->next = root->left;
                    prev = root->left;
                }
               
                if (root->right){
                    if (prev)
                         prev->next = root->right;
                    prev = root->right;
                }
            }
            root = next;
        }
    }
};

2. Recursive 上级调试

class Solution {
public:
    void connect(TreeLinkNode *root) {
         if (root==NULL) return;
        
         TreeLinkNode dummy(-1);
         for (TreeLinkNode *cur = root, *prev=&dummy; cur; cur = cur->next)
         {
             if (cur->left!=NULL){
                 prev->next = cur->left;
                 prev = prev->next;
             }
            
             if (cur->right!=NULL){
                 prev->next = cur->right;
                 prev = prev->next;
             }
         }
         connect(dummy.next);
    
    }
};