LeetCode——Convert Sorted List to Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/pickless/article/details/11885291

本文介绍了一种将已排序的单链表转换为高度平衡的二叉搜索树的方法，通过逐步遍历链表并构建树的节点来实现。

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

采用了一种很笨的方法。

每次都遍历到中间那个node，然后分左右。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    TreeNode *build(ListNode *head, int l, int r) {
        TreeNode *root;
        if (l > r) {
            root = NULL;
        }
        else {
            ListNode *ptr = head;
            int mid = (l + r) / 2;
            for (int i = 0; i < mid; i++) {
                ptr = ptr->next;
            }
            root = new TreeNode(ptr->val);
            root->left = build(head, l, mid - 1);
            root->right = build(head, mid + 1, r);
        }
        return root;
    }
    
public:
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL) {
            return NULL;
        }
        ListNode *ptr = head;
        int n = 0;
        while (ptr->next != NULL) {
            ptr = ptr->next;
            n++;
        }
        return build(head, 0, n);
    }
};