题目

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:

  1. Given the sorted linked list: [-10,-3,0,5,9],
  3. One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
  5.       0
  6.      / \
  7.    -3   9
  8.    /   /
  9.  -10  5

题意

将一个有序链表转换成左右子树高度差不超过1的平衡二叉查找树。

思路

比较简单的方法是，将链表中的值存入数组中，接下来与 108. Convert Sorted Array to Binary Search Tree 一样进行二分递归。

直接用快慢指针可以一次遍历找到链表中的中位数：初始时快慢指针同时指向头结点，每次移动慢指针走1步、快指针走2步，当快指针无法继续走时慢指针正好指在中位数处。每次找到当前链表的中位数作为当前子树的根，以中位数为中心划分出左右链表，递归生成左右子树。

模拟中序遍历：很玄妙，利用了二叉查找树的中序遍历是递增序列的性质，具体还是看官方解答 - Approach 3: Inorder Simulation。

代码实现

Java

快慢指针

  1. class Solution {
  2.     public TreeNode sortedListToBST(ListNode head) {
  3.         if (head == null) {
  4.             return null;
  5.         }
  7.         ListNode mid = findMid(head);
  9.         TreeNode x = new TreeNode(mid.val);
  12.         x.left = mid == head ? null : sortedListToBST(head);
  13.         x.right = sortedListToBST(mid.next);
  15.         return x;
  16.     }
  18.     private ListNode findMid(ListNode head) {
  19.         ListNode pre = null;
  20.         ListNode slow = head, fast = head;
  22.         while (fast.next != null && fast.next.next != null) {
  23.             pre = slow;
  24.             slow = slow.next;
  25.             fast = fast.next.next;
  26.         }
  28.         if (pre != null) {
  29.             pre.next = null;
  30.         }
  32.         return slow;
  33.     }
  34. }

模拟中序遍历

  1. class Solution {
  2.     private ListNode head;
  4.     public TreeNode sortedListToBST(ListNode head) {
  5.         this.head = head;
  7.         // 求出链表长度
  8.         int len = 0;
  9.         ListNode p = head;
  10.         while (p != null) {
  11.             len++;
  12.             p = p.next;
  13.         }
  15.         return sortedListToBST(0, len - 1);
  16.     }
  18.     private TreeNode sortedListToBST(int left, int right) {
  19.         if (left > right) {
  20.             return null;
  21.         }
  23.         int mid = (left + right) / 2;
  25.         TreeNode leftChild = sortedListToBST(left, mid - 1);
  26.         TreeNode root = new TreeNode(head.val);
  27.         root.left = leftChild;
  28.         head = head.next;
  29.         root.right = sortedListToBST(mid + 1, right);
  31.         return root;
  32.     }
  33. }

JavaScript

  1. /**
  2.  * @param {ListNode} head
  3.  * @return {TreeNode}
  4.  */
  5. var sortedListToBST = function (head) {
  6.   const nums = []
  7.   while (head) {
  8.     nums.push(head.val)
  9.     head = head.next
  10.   }
  12.   return dfs(nums, 0, nums.length - 1)
  13. }
  15. var dfs = function (nums, left, right) {
  16.   if (left > right) return null
  18.   const mid = Math.trunc((right - left) / 2) + left
  19.   const root = new TreeNode(nums[mid])
  20.   root.left = dfs(nums, left, mid - 1)
  21.   root.right = dfs(nums, mid + 1, right)
  22.   return root
  23. }