链表 有序链表转换二叉搜索树

题目

给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

示例:

给定的有序链表： [-10, -3, 0, 5, 9],

一个可能的答案是：[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树：

      0
     / \
   -3   9
   /   /
 -10  5

解题

/**
 * 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 && fastPtr.next != null) {
      prevPtr = slowPtr;
      slowPtr = slowPtr.next;
      fastPtr = fastPtr.next.next;
    }
    // Handling the case when slowPtr was equal to head.
    if (prevPtr != null) {
      prevPtr.next = 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(mid.next);
    return node;
  }
}