5.1.10 Populating Next Right Pointers in Each Node II

Notes:
	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. recursive solution.
	*/
	/**
	* 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) {}
	* };
	*/

void connect_2(TreeLinkNode *root) {
        if (root == NULL) return;
        queue<TreeLinkNode *> q;
        q.push(root);
        q.push(NULL);
        TreeLinkNode *last = NULL;
        TreeLinkNode dummy(-1);
        TreeLinkNode *pre = &dummy;
        while (!q.empty()) {
            TreeLinkNode *node = q.front();
            q.pop();
            if (node == NULL) {
                if (dummy.next) q.push(NULL);
                pre = &dummy;
                pre->next = NULL;
            } else {
                pre->next = node;
                pre = pre->next;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
    }