Java implementation LeetCode level binary tree traversal 102

102. The binary tree hierarchy traversal

Given a binary tree, the node returns its value 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]
]

 
class Solution {
   public List<List<Integer>> levelOrder(TreeNode root) {
    if(root == null)
        return new ArrayList<>();
    List<List<Integer>> res = new ArrayList<>();
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.add(root);
    while(!queue.isEmpty()){
        int count = queue.size();
        List<Integer> list = new ArrayList<Integer>();
        while(count > 0){
            TreeNode node = queue.poll();
            list.add(node.val);
            if(node.left != null)
                queue.add(node.left);
            if(node.right != null)
                queue.add(node.right);
            count--;
        }
        res.add(list);
    }
    return res;
}
}