Convert Sorted Array to Binary Search Tree
Explore how to transform a sorted array into a height-balanced binary search tree. This lesson helps you understand tree depth-first search principles and build balanced BSTs by ensuring subtree height differences stay within one, supporting multiple valid solutions.
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Statement
Given an array of integers, nums, sorted in ascending order, your task is to construct a height-balanced binary search tree (BST) from this array.
In a height-balanced BST, the difference of heights of the left subtree and right subtree of any node is not more than 1.
Note: There can be multiple valid BSTs for a given input.
Constraints:
-
nums.length -
nums[i] numsis sorted in strictly ascending order.
Examples
Understand the problem
Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps us to check if you’re solving the correct problem:
Convert Sorted Array to a Binary Search Tree
Select all valid BSTs that can be created with the given sorted array:
[5, 10, 15, 20] Multi-select
5
/ \
15 10
\
20
10
/ \
5 15
\
20
15
/ \
5 20
/
10
15
/ \
10 20
/
5
Figure it out!
We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.
Try it yourself
Implement your solution in the following coding playground.
import java.util.*;import ds_v1.BinaryTree.TreeNode;// Definiton of a binary tree node class// class TreeNode<T> {// T data;// TreeNode<T> left;// TreeNode<T> right;// TreeNode(T data) {// this.data = data;// this.left = null;// this.right = null;// }// }public class Solution {public static TreeNode<Integer> sortedArrayToBST(int[] nums) {// Replace this placeholder return statement with your codereturn new TreeNode<Integer>(-1);}}