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;
}
}