[LeetCode] 109. 有序链表转换二叉搜索树

题目链接 : https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree/

题目描述:

给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

给定的有序链表： [-10, -3, 0, 5, 9],

一个可能的答案是：[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树：

      0
     /    -3   9
   /   /
 -10  5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/convert-sorted-list-to-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路:

与上一题108. 将有序数组转换为二叉搜索树,还是找中点

但是这个是链表找中点,所以我们用快慢指针!

代码:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedListToBST(self, head: ListNode) -> TreeNode:
        def findmid(head, tail):
            slow = head
            fast = head
            while fast != tail and fast.next!= tail :
                slow = slow.next
                fast = fast.next.next
            return slow

        def helper(head, tail):
            if  head == tail: return
            node = findmid(head, tail)
            root = TreeNode(node.val)
            root.left = helper(head, node)
            root.right = helper(node.next, tail)
            return root

        return helper(head, None)
            

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 TreeNode sortedListToBST(ListNode head) {
        if (head == null) return null;
        return helper(head, null);

    }

    private TreeNode helper(ListNode head, ListNode tail) {
        if (head == tail) return null;
        // mid
        ListNode slow = head;
        ListNode fast = head;
        while (fast != tail && fast.next != tail) {
            slow = slow.next;
            fast = fast.next.next;
        }
        TreeNode root = new TreeNode(slow.val);
        root.left = helper(head, slow);
        root.right = helper(slow.next, tail);
        return root;
    }
}

原文地址：https://www.cnblogs.com/powercai/p/11104608.html