LC109.java

/*
109. 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.

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
 */
 
 /**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private ListNode head;
    
     private int findSize(ListNode head) 
     {
        ListNode ptr = head;
        int c = 0;
        while (ptr != null)
        {
            ptr = ptr.next;  
            c += 1;
        }
        return c;
    }
    
    private TreeNode convertListToBST(int l, int r)
    {
        // Invalid case
        if (l > r) {
        return null;
        }
        
        int mid = (l + r) / 2;

        // First step of simulated inorder traversal. Recursively form
        // the left half
        TreeNode left = this.convertListToBST(l, mid - 1);

        // Once left half is traversed, process the current node
        TreeNode node = new TreeNode(this.head.val);
        node.left = left;

        // Maintain the invariance mentioned in the algorithm
        this.head = this.head.next;

        // Recurse on the right hand side and form BST out of them
        node.right = this.convertListToBST(mid + 1, r);
        return node;
    }
    
    public TreeNode sortedListToBST(ListNode head) 
    {
         // Get the size of the linked list first
        int size = this.findSize(head);

        this.head = head;

        // Form the BST now that we know the size
        return convertListToBST(0, size - 1);
    }
}