Convert Sorted List to Binary Search Tree

This is No.109 in leetcode.

Below is the 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; } }
class Solution {

  private ListNode findMiddleElement(ListNode head) {

    // The pointer used to disconnect the left half from the mid node.
    ListNode prevPtr = null;
    ListNode slowPtr = head;
    ListNode fastPtr = head;

    // Iterate until fastPr doesn't reach the end of the linked list.
    while (fastPtr != null && != null) {
      prevPtr = slowPtr;
      slowPtr =;
      fastPtr =;

    // Handling the case when slowPtr was equal to head.
    if (prevPtr != null) { = null;

    return slowPtr;

  public TreeNode sortedListToBST(ListNode head) {

    // If the head doesn't exist, then the linked list is empty
    if (head == null) {
      return null;

    // Find the middle element for the list.
    ListNode mid = this.findMiddleElement(head);

    // The mid becomes the root of the BST.
    TreeNode node = new TreeNode(mid.val);

    // Base case when there is just one element in the linked list
    if (head == mid) {
      return node;

    // Recursively form balanced BSTs using the left and right halves of the original list.
    node.left = this.sortedListToBST(head);
    node.right = this.sortedListToBST(;
    return node;

To solve this problem, I think here are some points I should remember:

  1. How to get the middle node of a linked list?
    Use the slow&fast points. The slow node move one node in a time, and the fast node move two nodes in a time. The final slow node will be the middle one.
  2. What it the height of BST?
    The height of a binary tree is the number of edges between the tree’s root and its furthest leaf.
  3. How to cur off the left halve of linked list?
    Use another node and set the next to null.

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