109. Convert Sorted List to Binary Search Tree
Tags: ‘Linked List’, ‘Depth-first Search’
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 of every node 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
Solution
/**
* 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; }
* }
*/
// Refer to leetcode solution 1
// 100% 92%%
class Solution {
public TreeNode sortedListToBST(ListNode head) {
if (head == null) return null;
ListNode mid = findMiddleNode(head);
TreeNode node = new TreeNode(mid.val);
if (head == mid) return node; // base case, only 1
node.left = sortedListToBST(head);
node.right = sortedListToBST(mid.next);
return node;
}
private ListNode findMiddleNode(ListNode head) {
ListNode prev = null, slow = head, fast = head;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
if (prev != null) { // left part not empty, cut left part off
prev.next = null;
}
return slow;
}
}