Convert Sorted Array to Binary Search Tree
Explore how to convert a sorted array into a height-balanced binary search tree by understanding the depth-first search traversal and balancing conditions. Learn to build valid BSTs that ensure the height difference between subtrees is minimal, improving tree performance and structure.
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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.
// Definition for a binary tree node// template<class T>// class TreeNode {// public:// T data;// TreeNode<T>* left;// TreeNode<T>* right;// TreeNode(const T data) : data(data), left(nullptr), right(nullptr) {}// };TreeNode<int>* SortedArrayToBst(std::vector<int>& nums) {// Replace this placeholder return statement with your codereturn new TreeNode<int>(-1);}