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,null,null,15,7],
    3
   / 
  9  20
    /  
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

难度：medium

题目：给定二叉树，返回其层次遍历结点值。

思路：队列(FIFO)

Runtime: 1 ms, faster than 86.63% of Java online submissions for Binary Tree Level Order Traversal.
Memory Usage: 37.5 MB, less than 100.00% of Java online submissions for Binary Tree Level Order Traversal.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (null == root) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int currentLevelCount = 1, nextLevelCount = 0;
        List<Integer> currentLevelElements = new ArrayList<>();
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            currentLevelElements.add(node.val);
            if (node.left != null) {
                queue.add(node.left);
                nextLevelCount++;
            }
            if (node.right != null) {
                queue.add(node.right);
                nextLevelCount++;
            }
            currentLevelCount--;
            if (0 == currentLevelCount) {
                result.add(currentLevelElements);
                currentLevelElements = new ArrayList<>();
                currentLevelCount = nextLevelCount;
                nextLevelCount = 0;
            }
        }
        
        return result;
    }
}