LeetCode Populating Next Right Pointers in Each Node II

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

因此，对一个节点需要向右找到第一个节点。

对于left，如果right不存在，就在father的next节点去找left/right，依次找下去。

对于right，直接在father的next节点开始找。

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeLinkNode node;
    public void connect(TreeLinkNode root) {
        if (root==null) {
            return;
        }

        TreeLinkNode left = root.left;
        TreeLinkNode right = root.right;

        if (left != null) {
            left.next = right;
            if (right == null) {
                node = root.next;
                while (node != null && left.next == null) {
                    left.next = node.left;
                    if (left.next == null) {
                        left.next = node.right;
                    }
                    node = node.next;
                }
            }
        }

        if (right != null) {
            node = root.next;
            while (node != null && right.next == null) {
                right.next = node.left;
                if (right.next == null) {
                    right.next = node.right;
                }
                node = node.next;
            }
        }

        connect(right);
        connect(left);
    }
}