0109. Convert Sorted List to Binary Search Tree (M)

Convert Sorted List to Binary Search Tree (M)

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:

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

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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。

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代码实现

Java

快慢指针

class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode mid = findMid(head);

        TreeNode x = new TreeNode(mid.val);


        x.left = mid == head ? null : sortedListToBST(head);
        x.right = sortedListToBST(mid.next);

        return x;
    }

    private ListNode findMid(ListNode head) {
        ListNode pre = null;
        ListNode slow = head, fast = head;

        while (fast.next != null && fast.next.next != null) {
            pre = slow;
            slow = slow.next;
            fast = fast.next.next;
        }

        if (pre != null) {
            pre.next = null;
        }

        return slow;
    }
}

模拟中序遍历

class Solution {
    private ListNode head;

    public TreeNode sortedListToBST(ListNode head) {
        this.head = head;
        
        // 求出链表长度
        int len = 0;
        ListNode p = head;
        while (p != null) {
            len++;
            p = p.next;
        }

        return sortedListToBST(0, len - 1);
    }

    private TreeNode sortedListToBST(int left, int right) {
        if (left > right) {
            return null;
        }

        int mid = (left + right) / 2;

        TreeNode leftChild = sortedListToBST(left, mid - 1);
        TreeNode root = new TreeNode(head.val);
        root.left = leftChild;
        head = head.next;
        root.right = sortedListToBST(mid + 1, right);

        return root;
    }
}

JavaScript

/**
 * @param {ListNode} head
 * @return {TreeNode}
 */
var sortedListToBST = function (head) {
  const nums = []
  while (head) {
    nums.push(head.val)
    head = head.next
  }

  return dfs(nums, 0, nums.length - 1)
}

var dfs = function (nums, left, right) {
  if (left > right) return null

  const mid = Math.trunc((right - left) / 2) + left
  const root = new TreeNode(nums[mid])
  root.left = dfs(nums, left, mid - 1)
  root.right = dfs(nums, mid + 1, right)
  return root
}