package ctci.chapterfour

// 4.2 - page 109
// Given a sorted (increasing order) array with unique integer elements, write an algorithm in Kotlin to create a binary
// search tree with minimal height.

fun main() {

    val arr = intArrayOf(1, 2, 3, 4, 5, 6, 7, 8, 9)
    val root = createMinimalBST(arr, 0, arr.size - 1)
    inOrderTraversal(root)
}

//This function uses a recursive approach to traverse the tree in-order, first traversing the left subtree, then
// visiting the current node, and finally traversing the right subtree.
fun inOrderTraversal(root: TreeNode?) {
    if (root == null) return
    inOrderTraversal(root.left)
    println(root.value)
    inOrderTraversal(root.right)
}

//One way to create a binary search tree with minimal height from a sorted array is to use a technique called
// "divide and conquer". The idea is to divide the array in half, take the middle element as the root of the tree,
// and recursively build the left and right subtrees using the elements to the left and right of the middle
// element respectively.

//This function takes an array of integers, and the start and end indices of the subarray to be used as input. It
// calculates the middle index of the subarray, creates a new TreeNode with the value at that index, and recursively
// calls itself to build the left and right subtrees of the node, passing the appropriate subarrays as input.

class TreeNode(val value: Int, var left: TreeNode? = null, var right: TreeNode? = null)

fun createMinimalBST(arr: IntArray, start: Int, end: Int): TreeNode? {
    if (start > end) return null

    val mid = (start + end) / 2
    val node = TreeNode(arr[mid])
    node.left = createMinimalBST(arr, start, mid - 1)
    node.right = createMinimalBST(arr, mid + 1, end)
    return node
}

//The time complexity of this algorithm is O(n), where n is the number of elements in the array, as it visits each
// element of the array once. The space complexity is O(n) as well, as it recursively calls itself for each element
// in the array.
package ctci.chapterfour

// 4.2 - page 109
// Given a sorted (increasing order) array with unique integer elements, write an algorithm in Kotlin to create a binary
// search tree with minimal height.

fun main() {

    val arr = intArrayOf(1, 2, 3, 4, 5, 6, 7, 8, 9)
    val root = createMinimalBST(arr, 0, arr.size - 1)
    inOrderTraversal(root)
}

//This function uses a recursive approach to traverse the tree in-order, first traversing the left subtree, then
// visiting the current node, and finally traversing the right subtree.
fun inOrderTraversal(root: TreeNode?) {
    if (root == null) return
    inOrderTraversal(root.left)
    println(root.value)
    inOrderTraversal(root.right)
}

//One way to create a binary search tree with minimal height from a sorted array is to use a technique called
// "divide and conquer". The idea is to divide the array in half, take the middle element as the root of the tree,
// and recursively build the left and right subtrees using the elements to the left and right of the middle
// element respectively.

//This function takes an array of integers, and the start and end indices of the subarray to be used as input. It
// calculates the middle index of the subarray, creates a new TreeNode with the value at that index, and recursively
// calls itself to build the left and right subtrees of the node, passing the appropriate subarrays as input.

class TreeNode(val value: Int, var left: TreeNode? = null, var right: TreeNode? = null)

fun createMinimalBST(arr: IntArray, start: Int, end: Int): TreeNode? {
    if (start > end) return null

    val mid = (start + end) / 2
    val node = TreeNode(arr[mid])
    node.left = createMinimalBST(arr, start, mid - 1)
    node.right = createMinimalBST(arr, mid + 1, end)
    return node
}

//The time complexity of this algorithm is O(n), where n is the number of elements in the array, as it visits each
// element of the array once. The space complexity is O(n) as well, as it recursively calls itself for each element
// in the array.