109. Convert Sorted List to Binary Search Tree
Medium

  • Given the head of 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

示例

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2

Input: head = []
Output: []

Example 3

Input: head = [0]
Output: [0]

Example 4

Input: head = [1,3]
Output: [3,1]

Constraints:

  • The number of nodes in head is in the range [0, 2 * 104].
  • -10^5 <= Node.val <= 10^5

# Javascript Solution

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {ListNode} head
 * @return {TreeNode}
 */
var sortedListToBST = function (head) {
  const array = [];
  converListToArray(head, array);
  if (array.length < 1) return null;
  return convertArrayToTree(array);
};

const converListToArray = (head, array) => {
  while (head) {
    array.push(head.val);
    head = head.next;
  }
};

const convertArrayToTree = array => {
  if (array.length <= 0) {
    return null;
  } else {
    let mid = parseInt(array.length / 2);
    let node = new TreeNode(array[mid]);
    let leftArray = array.slice(0, mid);
    let rightArray = array.slice(mid + 1);
    node.left = convertArrayToTree(leftArray);
    node.right = convertArrayToTree(rightArray);
    return node;
  }
};