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

題意：116題的followup，二叉樹不在保證是一棵完整二叉樹，并且要求只用O(1)的空間復雜度。

思路：
非滿二叉樹用116題的寬度優先搜索的方法仍然能夠解決，但是空間復雜度是O(n)。
O(1)空間復雜度沒有想到合適的方法，查看了discuss的解法。
discuss排名最高的解法，思路是用三個指針cur、pre和head：cur指向當前層遍歷到的節點，根據cur的left和right，將下一層節點的next連接起來；pre指向下一層遍歷到的前一個節點，主要用于連接下一層當前節點和它的前一個節點；head用于記錄下一層的第一個節點，用于更新cur。

public void connect(TreeLinkNode root) {
    if (root == null) {
        return;
    }

    TreeLinkNode cur = root;
    TreeLinkNode head = null;
    TreeLinkNode pre = null;

    while (cur != null) {
        while (cur != null) {
            if (cur.left != null) {
                if (pre == null) {
                    head = cur.left;
                    pre = cur.left;
                } else {
                    pre.next = cur.left;
                    pre = pre.next;
                }
            }
            if (cur.right != null) {
                if (pre == null) {
                    head = cur.right;
                    pre = cur.right;
                } else {
                    pre.next = cur.right;
                    pre = pre.next;
                }
            }
            cur = cur.next;
        }

        cur = head;
        //bug 如果head不置為null，會導致死循環
        head = null;
        pre = null;
    }
}