LeetCode in Kotlin

1367. Linked List in Binary Tree

Medium

Given a binary tree root and a linked list with head as the first node.

Return True if all the elements in the linked list starting from the head correspond to some downward path connected in the binary tree otherwise return False.

In this context downward path means a path that starts at some node and goes downwards.

Example 1:

Input: head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]

Output: true

Explanation: Nodes in blue form a subpath in the binary Tree.

Example 2:

Input: head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]

Output: true

Example 3:

Input: head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]

Output: false

Explanation: There is no path in the binary tree that contains all the elements of the linked list from head.

Constraints:

• The number of nodes in the tree will be in the range [1, 2500].
• The number of nodes in the list will be in the range [1, 100].
• 1 <= Node.val <= 100 for each node in the linked list and binary tree.

Solution

import com_github_leetcode.ListNode
import com_github_leetcode.TreeNode

/*
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
/*
 * 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 isSubPath(head: ListNode?, root: TreeNode?): Boolean {
        return if (root == null) {
            false
        } else (
            doesRootHaveList(head, root) ||
                isSubPath(head, root.left) ||
                isSubPath(head, root.right)
            )
    }

    private fun doesRootHaveList(head: ListNode?, root: TreeNode?): Boolean {
        if (head == null) {
            return true
        }
        return if (root == null) {
            false
        } else (
            head.`val` == root.`val` &&
                (
                    doesRootHaveList(head.next, root.left) ||
                        doesRootHaveList(head.next, root.right)
                    )
            )
    }
}

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