Trees

BST Insert

Insert values into a binary search tree by comparing at each node.

Algorithm

The canonical tree is 4(2(1,3),6(5,7)), so this Kotlin DSA implementation can be compared directly with the rest of the DSA track.

binary search tree Values smaller than a node go left; larger values go right.

Visual walkthrough

BST insertion is a comparison path. The pinned tree 4(2(1,3),6(5,7)) is shown with the inserted value taking its sorted slot.

Step 1 - Start at root

For value 5, compare with 4 first; 5 is larger, so move right.

First comparison: 5 > 4, so the search for the insert slot goes right.insert 54compare26137

Step 2 - Take the left slot under 6

At 6, value 5 is smaller, so it becomes the left child.

Second comparison: 5 < 6, so the open left slot is used.426compare135new7

Step 3 - Canonical tree

The resulting tree is the pinned shape 4(2(1,3),6(5,7)).

Final BST after 5 is present under 6.4261357

Basic Implementation

basic.kt 
class Node(val value: Int, var left: Node? = null, var right: Node? = null)
fun render(node: Node?): String {
    if (node == null) return "_"
    if (node.left == null && node.right == null) return node.value.toString()
    return "${node.value}(${render(node.left)},${render(node.right)})"
}
fun sampleTree() = Node(4, Node(2, Node(1), Node(3)), Node(6, Node(5), Node(7)))
fun listString(values: List<Int>) = values.joinToString(", ", "[", "]")
fun insert(root: Node?, value: Int): Node { if (root == null) return Node(value); if (value < root.value) root.left = insert(root.left, value) else root.right = insert(root.right, value); return root }
fun main() { var root: Node? = null; for (value in listOf(4, 2, 6, 1, 3, 5, 7)) root = insert(root, value); println(render(root)) }

Complexity

  • Time: O(h) per insert
  • Space: O(n)

Implementation notes

  • Kotlin represents nodes with class Node(val value: Int, var left: Node? = null, var right: Node? = null), so the stored value is immutable but child links are nullable and mutable.
  • main starts with var root: Node? = null and assigns root = insert(root, value) for each value, allowing the first insert to replace the null root with a new node.
  • insert(root: Node?, value: Int): Node allocates Node(value) only when the current subtree root is null; otherwise it mutates root.left or root.right and returns the same root reference.
  • The comparison branch is if (value < root.value) ... else ..., so equal values would follow the right-child path in this implementation.
  • Recursive calls write the returned subtree back into the nullable child slot: root.left = insert(root.left, value) or root.right = insert(root.right, value).
  • The trace inserts 4, 2, 6, 1, 3, 5, 7, producing 4(2(1,3),6(5,7)); it also records a sorted-insert contrast 1(_,2(_,3(_,4))) with height 4.
  • render(root) prints _ for null children and println(render(root)) outputs 4(2(1,3),6(5,7)).