Leetcode: Convert sorted list to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

Difficulty 70, similar to the convert sorted array to binary search tree problem. The practice of this question is that we need to use the dot to make the root of each layer of tree. The problem is that, for a linked list, we cannot access its intermediate elements at constant time. So my idea is to spend O (n) Time First saving the elements in the linked list to an array or arraylist, so that there is a sorted array or arraylist, we can easily recursively create a height balanced BST.

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { val = x; next = null; } 7  * } 8  */ 9 /**10  * Definition for binary tree11  * public class TreeNode {12  *     int val;13  *     TreeNode left;14  *     TreeNode right;15  *     TreeNode(int x) { val = x; }16  * }17  */18 public class Solution {19     public TreeNode sortedListToBST(ListNode head) {20         if (head == null) return null;21         ListNode dummy = new ListNode(-1);22         dummy.next = head;23         ListNode cursor = dummy;24         int count = 0;25         while (cursor.next != null) {26             count++;27             cursor = cursor.next;28         }29         int[] array = new int[count];30         cursor = dummy;31         for (int i = 0; i < count; i++) {32             array[i] = cursor.next.val;33             cursor = cursor.next;34         }35         TreeNode result = helper(array, 0, array.length - 1);36         return result;37     }38     39     public TreeNode helper(int[] array, int begin, int end) {40         if (begin > end) return null;41         if (begin == end) return new TreeNode(array[begin]);42         int mid = (begin + end) / 2;43         TreeNode root = new TreeNode(array[mid]);44         root.left = helper(array, begin, mid - 1);45         root.right = helper(array, mid + 1, end);46         return root;47     }48 }

Using arraylist:

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { val = x; next = null; } 7  * } 8  */ 9 /**10  * Definition for binary tree11  * public class TreeNode {12  *     int val;13  *     TreeNode left;14  *     TreeNode right;15  *     TreeNode(int x) { val = x; }16  * }17  */18 public class Solution {19     public TreeNode sortedListToBST(ListNode head) {20         if (head == null) return null;21         22         ArrayList<Integer> listNode = new ArrayList<Integer>();23         while(head != null){24             listNode.add(head.val);25             head = head.next;26         }27         int left = 0;28         int right= listNode.size()-1;29         return helper(listNode, left, right);30     }31     32     public TreeNode helper(ArrayList<Integer> listNode, int left, int right){33         if(left>right) return null;34         35         TreeNode node = new TreeNode(0);36         int mid = (left+right)/2;37         node.val = listNode.get(mid);38         node.left = helper(listNode, left, mid-1);39         node.right = helper(listNode, mid+1, right);            40         return node;41     }42 }

 

Leetcode: Convert sorted list to Binary Search Tree