109 Convert Sorted List to Binary Search Tree

109. Convert Sorted List to Binary Search Tree

1. Question

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees ofeverynode never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

2. Implementation

(1) Recursion

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

    public TreeNode convertToBST(ListNode start, ListNode end) {
        if (start == null || start == end) {
            return null;
        }

        ListNode fast = start;
        ListNode slow = start;

        while (fast.next != end && fast.next.next != end) {
            fast = fast.next.next;
            slow = slow.next;
        }

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

3. Time & Space Complexity

Recursion: 时间复杂度O(nlogn), 空间复杂度O(logn)