LintCode242. 将二叉树按照层级转化为链表

描述
中文
给一棵二叉树，设计一个算法为每一层的节点建立一个链表。也就是说，如果一棵二叉树有 D 层，那么你需要创建 D 条链表。
样例
样例 1:

输入: {1,2,3,4}
输出: [1->null,2->3->null,4->null]
解释:
1
/
2 3
/
4
样例 2:

输入: {1,#,2,3}
输出: [1->null,2->null,3->null]
解释:
1

2
/
3

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param root the root of binary tree
     * @return a lists of linked list
     */
    public List binaryTreeToLists(TreeNode root) {
        // Write your code here
         List> result = new ArrayList<>();
        List resultNode = new ArrayList<>();
        //判断根节点是否为空
        if (root == null) {
            return resultNode;
        }

        //定义一个队列,并添加根节点
        Queue queue = new LinkedList<>();
        queue.add(root);

        TreeNode node = null;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            result.add(level);
        }
        for (int i = 0; i < result.size(); i++) {
            ListNode temp = new ListNode(result.get(i).get(0));
            ListNode head = temp;
            for (int j = 1; j < result.get(i).size(); j++) {
                temp.next = new ListNode(result.get(i).get(j));
                temp = temp.next;
            }
            resultNode.add(head);
        }
        return resultNode;
    }
}