Binary Tree Level Order Traversal I,II

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]

ref http://www.cnblogs.com/springfor/p/3891391.html
知道用queue但是如何控制cur level和next level，ref给出了计数var， 标准的bfs做法

public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> res  = new ArrayList<ArrayList<Integer>>();
if(root==null) return res;
ArrayList<Integer> cl = new ArrayList<Integer>();
LinkedList<TreeNode> q  = new LinkedList<TreeNode>();
q.add(root);
int curNum=1, nextNum=0;
while(!q.isEmpty()){
TreeNode n = q.poll();
curNum--;
cl.add(n.val);
if(n.left!=null){
q.add(n.left);
nextNum++;
}
if(n.right!=null){
q.add(n.right);
nextNum++;
}
if(curNum==0){
res.add(cl);
cl = new ArrayList<Integer>();
curNum  =nextNum;
nextNum = 0;
}
}
return res;
}

Binary Tree Level Order Traversal II

res.add(0,cl);