109. Convert Sorted List to Binary Search Tree

[Tcdian]

109. Convert Sorted List to Binary Search Tree

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。

Example 1

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]

Note

• The numner of nodes in head is in the range [0, 2 * 10^4].
• -10^5 <= Node.val <= 10^5

[Tcdian]

Solution

• 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 values = getValues(head);
    return buildBST(0, values.length - 1);

    function getValues(head) {
        const result = [];
        while(head !== null) {
            result.push(head.val);
            head = head.next;
        }
        return result;
    }
    function buildBST(start, end) {
        if (start > end) {
            return null;
        }
        const mid = (start + end) >> 1;
        const root = new TreeNode(values[mid]);
        root.left = buildBST(start, mid - 1);
        root.right = buildBST(mid + 1, end);
        return root;
    }
};

• TypeScript Solution

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function sortedListToBST(head: ListNode | null): TreeNode | null {
    const values = getValues(head);
    return buildBST(0, values.length - 1);

    function getValues(head: ListNode | null): number[] {
        const result: number[] = [];
        while(head !== null) {
            result.push(head.val);
            head = head.next;
        }
        return result;
    }
    function buildBST(start: number, end: number): TreeNode | null {
        if (start > end) {
            return null;
        }
        const mid = (start + end) >> 1;
        const root = new TreeNode(values[mid]);
        root.left = buildBST(start, mid - 1);
        root.right = buildBST(mid + 1, end);
        return root;
    }
};