[LeetCode] Populating Next RIght Pointer 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

Key: consider the case when the next pointer has no child. Must find the following next pointers until one has a child. Populate from the right.

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(root==NULL || (root->left==NULL && root->right==NULL))
return;

if(root->left!=NULL && root->right!=NULL)
root->left->next=root->right;

TreeLinkNode *rootLastChild;
if(root->right!=NULL)
rootLastChild=root->right;
else
rootLastChild=root->left;

TreeLinkNode *nextChild=NULL;

TreeLinkNode *rootNext=root->next;
while(rootNext!=NULL)
{
if(rootNext->left!=NULL)
nextChild=rootNext->left;
else if(rootNext->right!=NULL)
nextChild=rootNext->right;

if(nextChild != NULL)
break;

rootNext=rootNext->next;
}

if(nextChild != NULL)
{
rootLastChild->next=nextChild;
}

connect(root->right);
connect(root->left);
}
};