(LeetCode) 108. Convert Sorted Array to Binary Search Tree

108. Convert Sorted Array to Binary Search Tree

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Trees
• 난이도 : Easy

Problem

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

Example 1

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,3] and [3,1] are both a height-balanced BSTs.

Constraints

• 1 <= nums.length <= 10^4
• -10^4 <= nums[i] <= 10^4
• nums is sorted in a strictly increasing order.

Solution

Exapnder

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun sortedArrayToBST(nums: IntArray): TreeNode? {
        return recursive(nums, 0, nums.size - 1)
    }

    fun recursive(nums: IntArray, low: Int, high: Int): TreeNode? {
        if (low > high) {
            return null
        }
        var mid = low + (high - low) / 2

        return TreeNode(nums[mid]).apply {
            left = recursive(nums, low, mid - 1)
            right = recursive(nums, mid + 1, high)
        }
    }
}

Point of Thinking

• 오름차순 정렬이니, 주어진 배열의 정중앙에 위치한 값이 root 노드이다.
• root노드를 기준으로 좌측과 우측에 대해서 재귀 호출로 정렬하면 Accepted.