/*
This implementation demonstrates how
to efficiently construct a min height
binary search tree based on a sorted
array of ints.
In a binary search tree, for ever node v:
- Elements in left subtree rooted at v
are less than element stored at v.
- Elements in right subtree rooted at v
are greater than or equal to one at v.
Let n be the number of nodes in the
binary search tree.
Time complexity: O(n)
Space complexity: O(log2(n))
*/
public class SortedArrayToBST {
BTNode BSTRoot;
public static void main(String[] args) {
SortedArrayToBST application = new SortedArrayToBST();
int[] nums = { -10, -3, 0, 5, 9 };
application.BSTRoot = application.convertArrayToBST(nums);
/*
* The below bst is created:
* .. 0
* ./..
* -10.. 5
* ....
* .-3 .. 9
*/
System.out.println(application.BSTRoot.val); // 0
System.out.println(application.BSTRoot.left.val); // -10
System.out.println(application.BSTRoot.left.right.val); // -3
System.out.println(application.BSTRoot.right.val); // 5
System.out.println(application.BSTRoot.right.right.val); // 9
}
public BTNode convertArrayToBST(int[] nums) {
if (nums == null || nums.length == 0) {
return null;
}
return constructBST(nums, 0, nums.length - 1);
}
private BTNode constructBST(int[] nums, int left, int right) {
if (left > right)
return null;
int middle = left + (right - left) / 2;
// Middle node is root for values between left and right
BTNode currentNode = new BTNode(nums[middle], null, null);
// Set left of current root
currentNode.left = constructBST(nums, left, middle - 1);
// Set right of current root
currentNode.right = constructBST(nums, middle + 1, right);
return currentNode;
}
// Class representing a binary tree node
// with pointers to value, left, right, and parent nodes
private class BTNode {
int val;
BTNode left;
BTNode right;
public BTNode(int val, BTNode left, BTNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}