【LeetCode OJ】Populating Next Right Pointers in Nod

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

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
		if(root == null)return;
		Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
		q.add(root);
		int n = 0;
		TreeLinkNode temp,head = new TreeLinkNode(0);
		while(!q.isEmpty()){
			temp = q.poll();
			if(temp.left != null){
				if(q.size() == Math.pow(2, n) - 1){
					++n;
					root.next = null;
					head = temp.left;
					head.next = temp.right;
					head = temp.right;
				}else{
					head.next = temp.left;
					temp.left.next = temp.right;
					head = temp.right;
				}
				q.add(temp.left);
				q.add(temp.right);
			}
		}        
    }
}