Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
Solution:
In order to avoid traversing an element more than once we keep a ListNode that its next element indicates which node is one that we have to process.
/**
* 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) {
int count = 0;
ListNode temp=head;
while(temp!=null)
{
count++;
temp= temp.next;
}
return sortedListToBST(head, count, new ListNode(0));
}
private TreeNode sortedListToBST(ListNode head, int elements, ListNode next)
{
if(elements == 0)
{
next.next = head;
return null;
}
TreeNode left = sortedListToBST(head, (elements)/2, next);
TreeNode root = new TreeNode(next.next.val);
next.next = next.next.next;
TreeNode right = sortedListToBST(next.next, elements-1-elements/2, next);
root.left = left;
root.right = right;
return root;
}
}
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