LeetCode in Kotlin

701. Insert into a Binary Search Tree

Medium

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Example 1:

Input: root = [4,2,7,1,3], val = 5

Output: [4,2,7,1,3,5]

Explanation: Another accepted tree is:

Example 2:

Input: root = [40,20,60,10,30,50,70], val = 25

Output: [40,20,60,10,30,50,70,null,null,25]

Example 3:

Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5

Output: [4,2,7,1,3,5]

Constraints:

• The number of nodes in the tree will be in the range [0, 104].
• -108 <= Node.val <= 108
• All the values Node.val are unique.
• -108 <= val <= 108
• It’s guaranteed that val does not exist in the original BST.

Solution

import com_github_leetcode.TreeNode

/*
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun insertIntoBST(
        root: TreeNode?,
        value: Int,
    ): TreeNode? {
        if (root == null) {
            return TreeNode(value)
        }
        when {
            root.value < value -> root.right = insertIntoBST(root.right, value)
            root.value > value -> root.left = insertIntoBST(root.left, value)
        }
        return root
    }

    private val TreeNode.value get() = `val`
}

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