Problem Description

You are given a binary tree and a linked list. Your task is to determine if the linked list represents a valid downward path somewhere in the binary tree.

A downward path in the binary tree means:

• It starts at any node in the tree (not necessarily the root)
• It moves only downward (from parent to child nodes)
• It can go either left or right at each step

The linked list matches a path if:

• Starting from the linked list's head, each value in the linked list corresponds to consecutive nodes along a downward path in the tree
• All values in the linked list must be matched in order

For example, if the linked list is 4 -> 2 -> 8 and the binary tree contains a path like 1 -> 4 -> 2 -> 8, then the linked list matches the subpath 4 -> 2 -> 8.

The function should return True if such a matching downward path exists anywhere in the binary tree, and False otherwise.

The solution uses a two-level approach:

1. The main function isSubPath checks every node in the tree as a potential starting point
2. For each node, it calls a helper function dfs that checks if the linked list matches a path starting from that specific node
3. The dfs function recursively compares the linked list values with the tree path, exploring both left and right children
4. If the entire linked list is matched (head becomes None), it returns True
5. If the tree path ends before the linked list is fully matched, or if values don't match, it returns False

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: A binary tree is a special type of graph where each node has at most two children and there are no cycles.

Is it a tree?

• Yes: The problem explicitly involves a binary tree structure where we need to traverse from parent nodes to child nodes.

DFS

• Yes: We arrive at DFS (Depth-First Search) as our algorithm choice.

Conclusion: The flowchart suggests using a DFS approach for this problem.

This makes perfect sense because:

1. We need to explore paths in a binary tree starting from various nodes
2. We need to check if a linked list matches any downward path in the tree
3. DFS allows us to explore each path completely before backtracking
4. The recursive nature of DFS naturally fits with following the linked list sequence while traversing down the tree
5. We can use DFS both to traverse the tree to find potential starting points and to verify if the linked list matches a path from each starting point

The solution implements DFS in two ways:

• The outer DFS traverses the entire tree to try each node as a potential starting point
• The inner DFS (the dfs helper function) checks if the linked list matches a path starting from a specific node

Intuition

The key insight is that we need to find if the linked list appears as a consecutive sequence somewhere in the tree, following a downward path. This immediately suggests we need to check multiple starting points - the linked list path could start at any node in the tree, not just the root.

Think of it like finding a substring in a string, but in a tree structure. Just as you would check every position in a string as a potential starting point for your substring, we need to check every node in the tree as a potential starting point for our linked list path.

This naturally leads to a two-level search strategy:

1. First level: Traverse the entire tree to try each node as a potential starting point. This ensures we don't miss any possible matching path that might start deep in the tree.

2. Second level: For each potential starting node, check if we can match the entire linked list by following a path downward from that node.

The second level check is straightforward - we compare values one by one:

• If the current tree node matches the current linked list node, we continue checking with the next linked list node and either the left or right child
• If we successfully match all linked list nodes (reach the end of the list), we've found our answer
• If values don't match or we run out of tree nodes before finishing the linked list, this path doesn't work

Why DFS? Because we need to explore complete paths. When checking if a linked list matches from a particular starting node, we need to go all the way down one path before trying another. BFS would jump between different paths at the same level, which doesn't help us verify consecutive sequences.

The recursive nature of DFS also elegantly handles the backtracking - if one path doesn't work (say going left), we automatically try the other path (going right) through the or operation in our recursive calls.

Learn more about Tree, Depth-First Search, Linked List and Binary Tree patterns.

Solution Approach

The solution implements a recursive approach with two functions working together:

Main Function: isSubPath

This function traverses the entire binary tree to try each node as a potential starting point for matching the linked list. For each node in the tree, it:

• First checks if the current node is None (base case - no more nodes to check)
• Calls the helper function dfs to check if the linked list matches starting from this node
• Recursively checks the left and right subtrees as potential starting points
• Returns True if any starting point results in a match

The function uses the logical or operator to combine three checks:

dfs(head, root) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)

Helper Function: dfs

This function verifies if the linked list matches a downward path starting from a specific tree node. The logic follows these steps:

1. Base Case - Success: If head is None, we've successfully matched the entire linked list, return True

2. Base Case - Failure: If root is None (ran out of tree nodes) or root.val != head.val (values don't match), this path doesn't work, return False

3. Recursive Case: If the current values match, continue checking:

   • Try matching the rest of the linked list (head.next) with the left child (root.left)
   • Try matching the rest of the linked list (head.next) with the right child (root.right)
   • Return True if either path works

The recursion naturally handles backtracking - if the left path doesn't work, it tries the right path automatically.

Time Complexity: O(N × M) where N is the number of nodes in the tree and M is the length of the linked list. In the worst case, we check M nodes for each of the N tree nodes.

Space Complexity: O(H) where H is the height of the tree, due to the recursive call stack. In the worst case (skewed tree), this could be O(N).

Example Walkthrough

Let's walk through a concrete example to illustrate how the solution works.

Given:

• Linked List: 4 -> 2 -> 8
• Binary Tree:

        1
       / \
      4   4
     /     \
    2       2
   /         \
  1           8

Step-by-step execution:

1. Start with root (1):

   • Call isSubPath(head=4, root=1)
   • First check: dfs(head=4, root=1)
     • Values don't match (4 ≠ 1), return False
   • Continue to left subtree: isSubPath(head=4, root.left=4)

2. Check left child of root (node with value 4):

   • Call isSubPath(head=4, root=4)
   • First check: dfs(head=4, root=4)
     • Values match (4 = 4)! Continue with head.next=2
     • Try left: dfs(head=2, root.left=2)
       • Values match (2 = 2)! Continue with head.next=8
       • Try left: dfs(head=8, root.left=1)
         • Values don't match (8 ≠ 1), return False
       • Try right: dfs(head=8, root.right=None)
         • root is None, return False
       • Both paths failed, return False
   • The dfs from this node failed, but we still check this node's children
   • Continue to left subtree: isSubPath(head=4, root.left=2)

3. Check node with value 2 (left subtree):

   • Call isSubPath(head=4, root=2)
   • First check: dfs(head=4, root=2)
     • Values don't match (4 ≠ 2), return False
   • Continue checking children (node with value 1)
   • These won't match either

4. Eventually check right child of root (the other node with value 4):

   • Call isSubPath(head=4, root=4)
   • First check: dfs(head=4, root=4)
     • Values match (4 = 4)! Continue with head.next=2
     • Try left: dfs(head=2, root.left=None)
       • root is None, return False
     • Try right: dfs(head=2, root.right=2)
       • Values match (2 = 2)! Continue with head.next=8
       • Try left: dfs(head=8, root.left=None)
         • root is None, return False
       • Try right: dfs(head=8, root.right=8)
         • Values match (8 = 8)! Continue with head.next=None
         • head is None - we've matched the entire list!
         • Return True

5. Result: The function returns True because we found the path 4 -> 2 -> 8 in the right subtree.

Key observations from this walkthrough:

• We check every node as a potential starting point
• When we find a matching value, we explore both left and right paths
• We only return True when we match the entire linked list
• The recursion handles backtracking automatically when a path fails

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × m) where n is the number of nodes in the binary tree and m is the length of the linked list.

The analysis breaks down as follows:

• The main function isSubPath visits each node in the binary tree at most once, giving us O(n) iterations
• At each tree node, we potentially call the helper function dfs which traverses the linked list to check for a matching path
• The dfs function can traverse up to m nodes in the linked list in the worst case
• Therefore, the overall time complexity is O(n × m)

Note: The reference answer simplifies this to O(n²) which assumes the worst case where m ≈ n (the linked list length is comparable to the tree size).

Space Complexity: O(h) where h is the height of the binary tree, which is O(n) in the worst case.

The space complexity comes from:

• The recursion stack for isSubPath which can go as deep as the height of the tree: O(h)
• The recursion stack for dfs which can add another O(m) in the worst case when checking a path
• Combined, this gives us O(h + m), but since h can be O(n) in a skewed tree and typically m ≤ n, the space complexity simplifies to O(n) in the worst case

The reference answer's O(n) space complexity aligns with this analysis for the worst-case scenario of an unbalanced tree.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusing Path Continuity Requirements

A critical mistake is misunderstanding that once we start matching the linked list with a tree path, we must continue matching consecutive nodes without "restarting" the linked list in the middle of the path.

Incorrect Approach:

def dfs(head, root):
    if head is None:
        return True
    if root is None:
        return False
  
    # WRONG: This allows restarting the linked list matching at any point
    if root.val == head.val:
        return dfs(head.next, root.left) or dfs(head.next, root.right)
    else:
        # This incorrectly tries to restart matching from the beginning
        return dfs(head, root.left) or dfs(head, root.right)

This faulty logic would incorrectly return True for cases where the linked list values appear in the tree but not as a consecutive downward path.

Correct Approach: The solution correctly separates the two concerns:

• isSubPath: Tries different starting points in the tree
• dfs/check_path_from_node: Once started, must match the entire linked list consecutively

2. Not Handling Edge Cases Properly

Common Edge Case Mistakes:

• Not checking if the tree is empty before processing
• Not handling when the linked list has only one node
• Forgetting that the path can start from ANY node, not just the root

Example of Missing Edge Case Handling:

def isSubPath(self, head, root):
    # WRONG: Doesn't handle empty tree case first
    return (check_path_from_node(head, root) or 
            self.isSubPath(head, root.left) or  # This would crash if root is None
            self.isSubPath(head, root.right))

Solution: Always check for None cases first:

if root is None:
    return False

3. Inefficient Repeated Traversals

While the current solution is correct, a common pitfall when trying to optimize is to accidentally break the logic. Some might try to combine the traversal and matching in one function, leading to incorrect results.

Tempting but Wrong Optimization:

def isSubPath(self, head, root, original_head=None):
    if original_head is None:
        original_head = head
  
    # Trying to do everything in one function often leads to bugs
    # Missing proper reset logic or incorrect state management

Better Optimization (if needed): Instead of trying to merge functions, consider using memoization or converting the linked list to an array first for easier pattern matching algorithms like KMP, though this adds space complexity.

4. Incorrect Short-Circuit Logic

Wrong Use of Logical Operators:

# WRONG: Using 'and' instead of 'or'
return check_path_from_node(head, root) and (
    self.isSubPath(head, root.left) or self.isSubPath(head, root.right))

This would require the path to exist from the current node AND also exist in a subtree, which is logically incorrect.

Correct Logic: Use or to check if the path exists in ANY of these locations:

• Starting from current node
• Anywhere in the left subtree
• Anywhere in the right subtree