LeetCode-in-Java

109. Convert Sorted List to Binary Search Tree

Medium

Given the head of 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 of every node never differ by more than 1.

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: []

Example 3:

Input: head = [0]

Output: [0]

Example 4:

Input: head = [1,3]

Output: [3,1]

Constraints:

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

Solution

import com_github_leetcode.ListNode;
import com_github_leetcode.TreeNode;

/*
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        // Empty list -> empty tree / sub-tree
        if (head == null) {
            return null;
        }
        // No next node -> this node will become leaf
        if (head.next == null) {
            TreeNode leaf = new TreeNode(head.val);
            leaf.left = null;
            leaf.right = null;
            return leaf;
        }
        ListNode slow = head;
        // Head-Start fast by 1 to get slow = mid -1
        ListNode fast = head.next.next;
        // Find the mid of list
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        // slow.next ->  mid = our "root"
        TreeNode root = new TreeNode(slow.next.val);
        // Right sub tree from mid - end
        root.right = sortedListToBST(slow.next.next);
        // Left sub tree from head - mid (chop slow.next)
        slow.next = null;
        root.left = sortedListToBST(head);
        return root;
    }
}

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