leetcode Convert Sorted List to Binary Search Tree题解

最新推荐文章于 2020-04-13 06:45:19 发布

原创最新推荐文章于 2020-04-13 06:45:19 发布 · 168 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

162 篇文章

订阅专栏

博客围绕将有序链表转换为平衡二叉树展开。题目要求把升序排列的单链表转为高度平衡的二叉树，即每个节点的左右子树深度差不超1。解题思路是取链表中间节点为根节点，前后部分分别作为左右子树递归处理，分四种情况考虑，并给出了Java代码。

题目描述：

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

中文理解：

给定一个有序链表，将该有序链表转换为平衡二叉树。

解题思路：

如果将有序链表转换为平衡二叉树，肯定是将链表的中间节点转换为二叉树的根节点，然后中间节点前后分别作为根节点的左右子树进行递归，不过要分链表长度为0,1,2，大于2的四种情况来考虑。具体参见代码。

代码（java）：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int getListLength(ListNode head){
        int res=0;
        while(head!=null){
            res++;
            head=head.next;
        }
        return res;
    }
    public TreeNode sortedListToBST(ListNode head) {
        TreeNode root=null;
        if(head==null)return root;
        if(getListLength(head)==1)root=new TreeNode(head.val);
        else if(getListLength(head)==2){
            int max=Math.max(head.val,head.next.val);
            int min=Math.min(head.val,head.next.val);
            root=new TreeNode(max);
            root.left=new TreeNode(min);
        }
        else if(getListLength(head)>2){
            ListNode fast=head,slow=head;
            while(fast!=null && fast.next!=null){
                fast=fast.next.next;
                slow=slow.next;
            }
            root=new TreeNode(slow.val);
            root.right=sortedListToBST(slow.next);
            ListNode p=head;
            while(p!=null && p.next!=slow){
                p=p.next;
            }
            if(p!=null)p.next=null;
            root.left=sortedListToBST(head);
        }
        return root;
    }
}