Populating Next Right Pointers in Each Node (LeetCode Tree)

Question: 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

Idea: Breadth-first search. If the node visited is not the last node in the level, connect it to the next node in the queue. Depth-first search method is the same, just remember the last node.

Time: O(n) Space: O(n) for breadth-first search, O(1) for depth-first search.

Breadth-first Search Code:

 public class Solution {  
   public void connect(TreeLinkNode root) {  
     if(root==null)   
       return;  
     Queue<TreeLinkNode> queue=new LinkedList<TreeLinkNode>();  
     queue.offer(root);  
     while(queue.isEmpty()==false)  
     {  
       int queueSize=queue.size();  
       for(int i=0;i<queueSize;i++)  
       {  
         TreeLinkNode cur=queue.poll();  
         if(i!=queueSize-1)  
         {  
           cur.next=queue.peek();  
         }  
         if(cur.left!=null)  
           queue.offer(cur.left);  
         if(cur.right!=null)  
           queue.offer(cur.right);  
       }  
     }  
   }  
 }  

Depth-first Search Code:

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