LeetCode [102] binary tree hierarchy traversal

Title:
Given a binary tree node to return its value in a hierarchical traversal. (Ie, layer by layer, from left to right to access all nodes).

For example:
Given a binary tree: [3,9,20, null, null, 15,7],

3

/
920
/
157
to return to its level through the results:

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

/**

• Results returned by the observation of a large collection can be nested up to a small set, it requires two sets;
• Queue sets a queue, to hold the elements of each layer, and set a variable COUNT int type, this layer has recorded several elements;
• Each traversed layer, this layer is put queue element to poll off and added to the collection; poll difference and pop, pop null Throws
• If the left node and a right node through the previous values ​​stored in the queue is determined to be blank the next layer is added to the queue elements (since only the binary tree left node and a right node);
• Cycle is repeated until the queue is empty.
  * /

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> resultList = new ArrayList<>();
        if(root == null)
            return resultList;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        //将每层的节点放入Queue
        while(!queue.isEmpty()){
            List<Integer> levelList = new ArrayList<>();
            int count = queue.size();
            for(int i=0;i<count;i++){
                TreeNode node =  queue.poll();
                levelList.add(node.val);
                if(node.left!=null){
                    queue.add(node.left);
                }
                if(node.right!=null){
                    queue.add(node.right);
                }
            }
            resultList.add(levelList);
        }
        return resultList;
    }