[LeetCode] Populating Next Right Pointers in Each Node II

Problem Statement (link):

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

Analysis:
The idea is similar to the previous problem I. The difference is that we couldn't simply connect nodes in the same layer easily. We need an extra pointer prev to record the previous visited node on the same layer. See the following iterative solution for details.

Code:

/**
 * 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) {
        while(root!=NULL) {
            TreeLinkNode *curr=root;
            TreeLinkNode *prev=NULL;

            while (curr!=NULL) { // traverse each layer
                if (curr->left!=NULL) {
                    if (prev!=NULL)
                        prev->next=curr->left;
                    prev=curr->left;
                }
                if (curr->right!=NULL) {
                    if (prev!=NULL) 
                        prev->next=curr->right;
                    prev=curr->right;
                }
                curr=curr->next;
            }

            // find starting node in next layer
            while(root!=NULL) {
                if (root->left!=NULL) {
                    root=root->left;
                    break;
                }
                if (root->right!=NULL) {
                    root=root->right;
                    break;
                }
                root=root->next;
            }
        }
    }
};