109. Convert Sorted List to Binary Search Tree

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109. Convert Sorted List to Binary Search Tree

Description

Given the head of a singly linked list where elements are sorted in ascending order , convert it to a height-balanced binary search tree.

Example 1:

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Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

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Input: head = []
Output: []

Constraints:

  • The number of nodes in head is in the range [0, 2 * 10^4].
  • -10^5 <= Node.val <= 10^5

Hints/Notes

  • 2024/07/07
  • linked list and binary search tree
  • No solution from 0x3F

Solution

Language: C++

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/**
 * 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:
    ListNode* cur;

    TreeNode* sortedListToBST(ListNode* head) {
        int len = 0;
        for (ListNode* p = head; p; p = p->next) {
            len++;
        }
        cur = head;
        return traverse(0, len - 1);
    }

    TreeNode* traverse(int start, int end) {
        if (start > end) {
            return nullptr;
        }
        int mid = (end - start) / 2 + start;
        TreeNode* l = traverse(start, mid - 1);
        TreeNode* root = new TreeNode(cur->val);
        cur = cur->next;
        TreeNode* r = traverse(mid + 1, end);
        root->left = l;
        root->right = r;
        return root;
    }
};