Total Accepted: 9476 Total Submissions: 35578
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; next = null; }
* }
*/
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
if (head == null) return null;
if (head.next == null) return new TreeNode(head.val);
int len = 0;
ListNode lo = head;
ListNode hi, mid;
// get length of list
while (lo != null) {
len++;
lo = lo.next;
}
// get the middle index
int midIndex = (len >> 1) - 1;
lo = head;
while (midIndex-- > 0) lo = lo.next;
// divide original list into 3 parts
mid = lo.next;
hi = mid.next;
lo.next = null;
// build BST recursively
TreeNode root = new TreeNode(mid.val);
root.left = sortedListToBST(head);
root.right = sortedListToBST(hi);
return root;
}
}
// use a end flag
public class Solution {
public TreeNode sortedListToBST(ListNode head) {
return getBST(head, null);
}
public TreeNode getBST(ListNode start, ListNode endFlag) {
if (start == endFlag) return null;
ListNode hi = start;
ListNode mid = start;
// get mid node with 2 pointers
while (hi != endFlag && hi.next != endFlag) {
mid = mid.next;
hi = hi.next.next;
}
TreeNode root = new TreeNode(mid.val);
root.left = getBST(start, mid);
root.right = getBST(mid.next, endFlag);
return root;
}
}