LeetCode in Kotlin

700. Search in a Binary Search Tree

Easy

You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:

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

Output: [2,1,3]

Example 2:

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

Output: []

Constraints:

• The number of nodes in the tree is in the range [1, 5000].
• 1 <= Node.val <= 107
• root is a binary search tree.
• 1 <= val <= 107

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
 * }
 */
@Suppress("NAME_SHADOWING")
class Solution {
    fun searchBST(root: TreeNode?, `val`: Int): TreeNode? {
        var root: TreeNode? = root
        while (root != null && root.`val` != `val`) {
            root = if (root.`val` > `val`) {
                root.left
            } else {
                root.right
            }
        }
        return root
    }
}

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