Populating Next Right Pointers in Each Node II

/**
 * 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) {
        TreeLinkNode* cur = root;
        while(cur) {
            TreeLinkNode* node = cur;
            TreeLinkNode* last = NULL;
            cur = NULL;
            while(node) {
                TreeLinkNode* left = node->left;
                TreeLinkNode* right = node->right;
                if(left || right) {
                    if(last) last->next = left ? left : right;
                    if(left) left->next = right;
                    if(!cur) cur = left ? left : right;
                    last = right ? right : left;
                }
                node = node->next;
            }
        }
    }
};

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. iterative way with CONSTANT extra space.
2. iterative way + queue. Contributed by SUN Mian(孙冕).
3. tail recursive solution.
*/