leetCode-109:Convert Sorted List to Binary Search Tree

问题描述

给定一个升序的链表，要求将其转换成高度平衡的搜索二叉树。题目链接：**点我**

样例输入输出

输入：1->2->3

输出：

2

/ \

1 3

输入：1

输出：1

问题解法

此题跟 LeetCode-108类似，只不过前面是数组，这次是链表。解法类似，先找到中间的节点，然后拆分左子树和右子树，依次进行递归构造。代码如下

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        return convert2BST(head, null);
    }

    private TreeNode convert2BST(ListNode left, ListNode right) {
        if (left == right) {
            return null;
        }

        ListNode fast = left;
        ListNode slow = left;
        while (fast != right && fast.next != right) {
            fast = fast.next.next;
            slow = slow.next;
        }

        TreeNode root = new TreeNode(slow.val);
        root.left = convert2BST(left, slow);
        root.right = convert2BST(slow.next, right);
        return root;
    }
}