LeetCode in Kotlin

993. Cousins in Binary Tree

Easy

Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3

Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4

Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3

Output: false

Constraints:

• The number of nodes in the tree is in the range [2, 100].
• 1 <= Node.val <= 100
• Each node has a unique value.
• x != y
• x and y are exist in the tree.

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 isCousins(root: TreeNode?, x: Int, y: Int): Boolean {
        return !isSiblings(root, x, y) && isSameLevels(root, x, y)
    }

    private fun isSameLevels(root: TreeNode?, x: Int, y: Int): Boolean {
        return findLevel(root, x, 0) == findLevel(root, y, 0)
    }

    private fun findLevel(root: TreeNode?, x: Int, level: Int): Int {
        if (root == null) {
            return -1
        }
        if (root.`val` == x) {
            return level
        }
        val leftLevel = findLevel(root.left, x, level + 1)
        return if (leftLevel == -1) {
            findLevel(root.right, x, level + 1)
        } else {
            leftLevel
        }
    }

    private fun isSiblings(root: TreeNode?, x: Int, y: Int): Boolean {
        if (root == null) {
            return false
        }
        // Check children first
        val leftSubTreeContainsCousins = isSiblings(root.left, x, y)
        val rightSubTreeContainsCousins = isSiblings(root.right, x, y)
        if (leftSubTreeContainsCousins || rightSubTreeContainsCousins) {
            return true
        }
        return if (root.left == null || root.right == null) {
            false
        } else root.left!!.`val` == x && root.right!!.`val` == y ||
            root.right!!.`val` == x && root.left!!.`val` == y
    }
}

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