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]
]

分析

这题与107. Binary Tree Level Order Traversal II一样，都是二叉树的层序遍历，区别是102是从顶至底，107是从底至顶。
借助队列用BFS即可，核心思想是记录每一层中有几个元素（n），元素出队时把孩子节点入队，出队的元素放入临时数组，出够n个以后，临时数组中的值就是当前level的元素，把临时数组放入结果数组。对于102，临时数组放入结果数组的末尾，对于107，需要把临时数组放入结果数组的开头，因此对于结果数组而言，使用LinkedList更好。

code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new LinkedList<>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> tmp = new ArrayList<>();
            int n = queue.size();
            while (n > 0) {
                TreeNode node = queue.poll();
                tmp.add(node.val);
                if (node.left != null) 
                    queue.offer(node.left);
                if (node.right != null)
                    queue.offer(node.right);
                n--;
            }
            res.add(tmp);
            // res.add(0, tmp); // for 107
        }
        return res;
    }
}