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1367. Linked List in Binary Tree

MediumOpen on LeetCodeProblem statement

Problem Statement

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:

Java

Source file
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    ListNode ll;

    public boolean isSubPath(ListNode head, TreeNode root) {
        if (root == null)
            return false;
        ll = head;
        return solve(head, root) || isSubPath(head, root.left) || isSubPath(head, root.right);
    }

    private boolean solve(ListNode head, TreeNode root) {
        if (head == null)
            return true;
        if (root == null)
            return false;
        // System.out.println(head.val + "\t" + root.val);
        if (root.val != head.val)
            return solve(ll, root.left) || solve(ll, root.right);
        return solve(head.next, root.left) || solve(head.next, root.right);
    }
}