109. 有序链表转换二叉搜索树

题意

题目链接

思路

递归

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if (head == nullptr)
            return nullptr;
        
        if (head->next == nullptr)
            return new TreeNode(head->val);
        
        ListNode *p = head;
        ListNode *q = head;
        ListNode *pre = nullptr;

        while (q && q->next)
        {
            pre = p;
            p = p->next;
            q = q->next->next;
        }
        pre->next = nullptr;

        TreeNode *root = new TreeNode(p->val);
        root->left = sortedListToBST(head);
        root->right = sortedListToBST(p->next);
        return root;
    }
};