Populating Next Right Pointers in Each Node II

题目
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

答案
level order traversal using the next pointer, update next pointers along the traversal

public class Solution {
    public void update_pointers(TreeLinkNode[] runners, TreeLinkNode ptr) {
        if(ptr != null) {
            if(runners[0] == null) runners[0] = ptr;
            else if(runners[1] == null) runners[0].next = runners[1] = ptr;
            else {
                runners[0] = runners[1].next = ptr;
                runners[1] = null;
            }
            if(runners[2] == null) runners[2] = ptr;
        }
    }
    
    public void connect(TreeLinkNode root) {
        TreeLinkNode curr;
        while(root != null) {
            curr = root;
            TreeLinkNode[] runners = {null, null, null};
            while(curr != null) {
                // Iterate through the level under curr
                update_pointers(runners, curr.left);
                update_pointers(runners, curr.right);                
                curr = curr.next;
            }
           
            root = runners[2];
        }
    }
}