109. Convert Sorted List to Binary Search Tree
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: []
Constraints:
- The number of nodes in head is in the range [0, 2 * 104].
- -105 <= Node.val <= 105
/**
* 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) {
if(head==null) {
return null;
}
return toBST(head,null);
}
public TreeNode toBST(ListNode head, ListNode tail){
ListNode slow = head;
ListNode fast = head;
if(head==tail){
return null;
}
while(fast!=tail&&fast.next!=tail){
fast = fast.next.next;
slow = slow.next;
}
TreeNode root = new TreeNode(slow.val);
root.left = toBST(head, slow);
root.right = toBST(slow.next, tail);
return root;
}
}