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:

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:

Input: head = [] Output: []

Constraints:

• The number of nodes in head is in the range [0, 2 * 104].
• -105 <= Node.val <= 105

Code

概念同 Convert Sorted Array to Binary Search Tree，只是從 array 換成 linked list，因此找中點要用 slow fast pointer。

概念也有點像 Sort List 中用到的 merge sort。

/**
 * 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) return nullptr;
        if(!head->next) {
            TreeNode* node = new TreeNode(head->val);
            return node;
        }
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* prev = head;
        while(fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        // slow is mid
        prev->next = nullptr;
        TreeNode* mid = new TreeNode(slow->val);
        mid->right = sortedListToBST(slow->next);
        mid->left = sortedListToBST(head);
        return mid;
    }
};

Source

• Convert Sorted List to Binary Search Tree - LeetCode