领扣LintCode算法问题答案-242. 将二叉树按照层级转化为链表242. 将二叉树按照层级转化为链表鸣谢

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我是靠谱客的博主大方大炮，这篇文章主要介绍领扣LintCode算法问题答案-242. 将二叉树按照层级转化为链表242. 将二叉树按照层级转化为链表鸣谢，现在分享给大家，希望可以做个参考。

领扣LintCode算法问题答案-242. 将二叉树按照层级转化为链表

242. 将二叉树按照层级转化为链表

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 1：

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

解释:

复制代码

样例 2：

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

解释:

复制代码

复制代码

/**
 * 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<ListNode> binaryTreeToLists(TreeNode root) {
        // Write your code here
        List<ListNode>  listNodes = new ArrayList<ListNode>();
        if (root != null) {
            Queue<TreeNode> nodes     = new LinkedList<TreeNode>();
            nodes.add(root);

ListNode head = new ListNode(0);
            while (!nodes.isEmpty()) {
                int size = nodes.size();
                ListNode lastNode = head;
                while (size-- > 0) {
                    TreeNode treeNode = nodes.poll();
                    lastNode.next = new ListNode(treeNode.val);
                    lastNode = lastNode.next;

if (treeNode.left != null) {
                        nodes.offer(treeNode.left);
                    }
                    if (treeNode.right != null) {
                        nodes.offer(treeNode.right);
                    }
                }
                listNodes.add(head.next);
            }
        }
        return listNodes;
    }
}

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/**
 * 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<ListNode> binaryTreeToLists(TreeNode root) {
        // Write your code here
        List<ListNode>  listNodes = new ArrayList<ListNode>();
        if (root != null) {
            Queue<TreeNode> nodes     = new LinkedList<TreeNode>();
            nodes.add(root);

            ListNode head = new ListNode(0);
            while (!nodes.isEmpty()) {
                int size = nodes.size();
                ListNode lastNode = head;
                while (size-- > 0) {
                    TreeNode treeNode = nodes.poll();
                    lastNode.next = new ListNode(treeNode.val);
                    lastNode = lastNode.next;

                    if (treeNode.left != null) {
                        nodes.offer(treeNode.left);
                    }
                    if (treeNode.right != null) {
                        nodes.offer(treeNode.right);
                    }
                }
                listNodes.add(head.next);
            }
        }
        return listNodes;
    }
}