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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
对于二叉树:
1
/ \
2 3
/
4
返回3条链表:
[
1->null,
2->3->null,
4->null
]
解题思路:
类似于Lintcode 69:二叉树的层次遍历,只不过换成链表存储。
/**
* 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> res = new ArrayList<>();
LinkedList<TreeNode> queue = new LinkedList<>();
if(root == null)
return res;
queue.offer(root);
while(!queue.isEmpty()){
int len = queue.size();
ListNode list = new ListNode(0);//链表的头部
ListNode p = list;//遍历链表的指针
while(len-- != 0){
//将TreeNode转换为ListNode,并连接到list链表上
TreeNode treeNode = queue.poll();
ListNode node = new ListNode(treeNode.val);
p.next = node;
p = p.next;
if(treeNode.left != null)
queue.offer(treeNode.left);
if(treeNode.right != null)
queue.offer(treeNode.right);
}
p.next = null;//结尾指针置为null
res.add(list.next);
}
return res;
}
}