109. Convert Sorted List to Binary Search Tree

109. Convert Sorted List to Binary Search Tree
(AC)

/**
 * 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; }
 * }
 */
  
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
      if(head == null) return null;
      return helper(head, null);
    }
  
    public TreeNode helper(ListNode head, ListNode tail){
      if(head == tail) return null;/////
    
      ListNode slow = head;
      ListNode fast = head;
      
      
      while(fast != tail && fast.next != tail){
        slow = slow.next;
        fast = fast.next.next;
      }
      
      TreeNode root = new TreeNode(slow.val);
      root.left = helper(head, slow);
      root.right = helper(slow.next, tail);
      return root;
    }
}










/// NullPointerException


class Solution {
    public TreeNode sortedListToBST(ListNode head) {
      // base cases 
      // size = 0;
      if(head == null) return null;
      // size = 1
      if(head.next == null){
        TreeNode root = new TreeNode(head.next.val);
        return root;
      }
      
      // size = 2
      if(head.next.next == null){
        TreeNode next = new TreeNode(head.next.val);
        TreeNode root = new TreeNode(head.val);
        
        root.right = next;
        return root;
      }
      
      // size = 3 
      if(head.next.next.next == null){
        TreeNode root = new TreeNode(head.next.val);
        TreeNode right = new TreeNode(head.next.next.val);
        TreeNode left = new TreeNode(head.val);
        
        root.left = left;
        root.right = right;
        
        return root;
      }
      
      // induction rules
      
      ListNode mid = findMid(head);
      ListNode rootNode = mid.next;
      ListNode nextNode = rootNode.next;
      
      mid.next = null;
      rootNode.next = null;
      
      TreeNode root = new TreeNode(rootNode.val);
      root.left = sortedListToBST(head);
      root.right = sortedListToBST(nextNode);
      
      return root;
    }
    
    private ListNode findMid(ListNode head){
      ListNode fast = head;
      ListNode slow = head;
      
      while(fast.next != null && fast.next.next != null){
        fast = fast.next.next;
        slow = slow.next;
      }
      return slow;
    }
}

 

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


先按照第一种方法做。 第二种方法不太靠谱。。。