LeetCode 117 - 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

Solution 1:
반복 버전.

public void connect(TreeLinkNode root) {
    if(root == null) return;
    TreeLinkNode p = root.next;
    while(p!=null) {
        if(p.left != null) {
            p = p.left; 
            break;
        }
        if(p.right != null) {
            p = p.right; 
            break;
        }
        p = p.next;
    }
    if(root.right != null) {
        root.right.next = p;
        p = root.right;
    } 
    if(root.left != null) {
        root.left.next = p;
    }
    connect(root.right);
    connect(root.left);
}

 
Solution 2:
비귀속 버전.

public void connect(TreeLinkNode root)  {
    if(root == null)  return;  
    TreeLinkNode lastHead = root;  
    TreeLinkNode curHead = null, curPrev = null;
    while(lastHead!=null)  {  
        while(lastHead != null) {
            if(lastHead.left!=null) {  
                if(curHead == null) {  
                    curHead = lastHead.left;  
                    curPrev = curHead;  
                } else {  
                    curPrev.next = lastHead.left;  
                    curPrev = curPrev.next;  
                }  
            }  
            if(lastHead.right!=null) {  
                if(curHead == null) {  
                    curHead = lastHead.right;  
                    curPrev = curHead;  
                } else {  
                    curPrev.next = lastHead.right;  
                    curPrev = curPrev.next;  
                }  
            }                  
            lastHead = lastHead.next; 
        }
        lastHead = curHead;  
        curHead = null;  
    }  
}