Populating Next Right Pointers in Each Node leetcode

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

node的left child 要指向right child, 如果node.next 是null的话, right child指向null, 如果不是null, right child则指向root.next的left child
每个节点访问一次 说是时间O(n) 空间O(1)

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            root.left.next = root.right;
        }
        if (root.right != null) {
            if (root.next != null) {
                root.right.next = root.next.left;
            } else {
                root.right.next = null;
            }
        }
        connect(root.left);
        connect(root.right);
    }
}