<p>Given a binary tree <code>root</code> and a&nbsp;linked list with&nbsp;<code>head</code>&nbsp;as the first node.&nbsp;</p>

<p>Return True if all the elements in the linked list starting from the <code>head</code> correspond to some <em>downward path</em> connected in the binary tree&nbsp;otherwise return False.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/sample_1_1720.png" style="width: 220px; height: 280px;" /></strong></p>

<pre>
<strong>Input:</strong> head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
<strong>Output:</strong> true
<strong>Explanation:</strong> Nodes in blue form a subpath in the binary Tree.  
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/02/12/sample_2_1720.png" style="width: 220px; height: 280px;" /></strong></p>

<pre>
<strong>Input:</strong> head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no path in the binary tree that contains all the elements of the linked list from <code>head</code>.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree will be in the range <code>[1, 2500]</code>.</li>
	<li>The number of nodes in the list will be in the range <code>[1, 100]</code>.</li>
	<li><code>1 &lt;= Node.val&nbsp;&lt;= 100</code>&nbsp;for each node in the linked list and binary tree.</li>
</ul>

class Solution:
    def isSubPath(self, head: Optional[ListNode], root: Optional[TreeNode]) -> bool:
        # Helper function to check if there's a matching path from this tree node downwards.
        def dfs(head: Optional[ListNode], node: Optional[TreeNode]) -> bool:
            if not head:  # If we've reached the end of the linked list, we found a match.
                return True
            if not node:  # If we've reached a null node in the tree, the path can't continue.
                return False
            if head.val != node.val:  # If the values don't match, return False.
                return False
            # Recur down the left and right subtrees to continue matching the next node in the list.
            return dfs(head.next, node.left) or dfs(head.next, node.right)
        
        # Main function to start the matching from any node in the tree.
        if not root:
            return False
        # Check if the current tree node starts a valid path or check its left and right children.
        return dfs(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)
101