What is the time complexity of Linked List in Binary Tree?

The optimal solution has a time complexity of O(N).

What data structures are used to solve Linked List in Binary Tree?

This problem uses Linked List, Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Linked List in Binary Tree?

Insight 1: We can traverse the binary tree and for each node, check if the linked list is a subpath starting at that node. Insight 2: A dynamic programming approach can be used to efficiently determine if a subpath exists. By working backwards from the end of the linked list, we can build up a table indicating whether a portion of the linked list exists starting at a given node in the tree.

Which companies ask LeetCode 1367. Linked List in Binary Tree in interviews?

This problem is frequently asked at SoundHound.

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Linked List in Binary Tree - LeetCode 1367 Solution

Linked List in Binary Tree - Complete Solution Guide

Linked List in Binary Tree is LeetCode problem 1367, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to determine whether a linked list's sequence of values exists as a downward path within a binary tree. A downward path means a sequence of connected nodes starting from any node in the tree and going down towards its descendants (children). We need to return `True` if such a path exists, and `False` otherwise. The linked list represents the target path, and the binary tree represents the search space. The values in the linked list and the tree nodes can range from 1 to 100. The size of the tree is up to 2500 nodes and the linked list can have up to 100 nodes.

Solution Approach

The provided code uses a dynamic programming approach to solve this problem. It first extracts the values from the linked list into a list called `list_vals`. Then, it performs a breadth-first search (BFS) on the binary tree to get a list of all tree nodes called `all_tree_nodes` and a map from each node to its index in the `all_tree_nodes` list called `node_to_idx`. The core of the algorithm is a dynamic programming table `dp` where `dp[i]` is true if the *last* value of `list_vals` can be found at the `i`th tree node in `all_tree_nodes`. The code iterates backward through the `list_vals`, using the `dp` table from the previous iteration to determine the values of `new_dp`. If `new_dp[j]` is true, this implies that a subpath starting at the node `all_tree_nodes[j]` can lead to the end of the linked list sequence. The backward iteration continues, and finally, if any entry in the `dp` table is true, it means there exists a subpath in the tree matching the linked list.

Step-by-Step Algorithm

Step 1: Convert the linked list into a list of integers `list_vals` for easier access.
Step 2: Traverse the binary tree using BFS and store all tree nodes in a list `all_tree_nodes`.
Step 3: Create a hash map `node_to_idx` to map each tree node to its index in the `all_tree_nodes` list.
Step 4: Initialize a DP table `dp` of size N (number of tree nodes) where `dp[j]` is true if the last element of the linked list `list_vals` is found at node `all_tree_nodes[j]`.
Step 5: Iterate backward from the second to last element of `list_vals` to the first, updating the DP table at each step.
Step 6: In each inner loop iteration, check if the current tree node's value matches the current element in `list_vals`. If it does, check if either its left or right child (if they exist) has a value of `True` in the previous DP table.
Step 7: After the outer loop finishes, check if any element in the DP table is `True`. If so, the linked list is a subpath of the binary tree. Return `True` if a subpath exists, `False` otherwise.

Key Insights

Insight 1: We can traverse the binary tree and for each node, check if the linked list is a subpath starting at that node.
Insight 2: A dynamic programming approach can be used to efficiently determine if a subpath exists. By working backwards from the end of the linked list, we can build up a table indicating whether a portion of the linked list exists starting at a given node in the tree.
Insight 3: Instead of recursively calling the subpath check function on each node, a more efficient approach pre-processes the tree into a list of all nodes and creates a map from node to index to avoid redundant traversals when using DP.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(N)

Topics

This problem involves: Linked List, Tree, Depth-First Search, Binary Tree.

Companies

Asked at: SoundHound.

Linked List in Binary Tree | LeetCode 1367

Medium

Linked List Tree Depth-First Search Binary Tree

Time: O(N)Space: O(N)

Solution Code

Available in:

/**
 * 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;
 *     }
 * }
 */
import java.util.*;

class Solution {
    public boolean isSubPath(ListNode head, TreeNode root) {
        if (root == null) {
            return false;
        }

        List<Integer> list_vals = new ArrayList<>();
        ListNode current_node = head;
        while (current_node != null) {
            list_vals.add(current_node.val);
            current_node = current_node.next;
        }

        int L = list_vals.size();

        List<TreeNode> all_tree_nodes = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            all_tree_nodes.add(node);
            if (node.left != null) {
                q.offer(node.left);
            }
            if (node.right != null) {
                q.offer(node.right);
            }
        }

        int N = all_tree_nodes.size();
        Map<TreeNode, Integer> node_to_idx = new HashMap<>();
        for (int i = 0; i < N; i++) {
            node_to_idx.put(all_tree_nodes.get(i), i);
        }

        boolean[] dp = new boolean[N];
        for (int j = 0; j < N; j++) {
            if (all_tree_nodes.get(j).val == list_vals.get(L - 1)) {
                dp[j] = true;
            }
        }

        for (int i = L - 2; i >= 0; i--) {
            boolean[] new_dp = new boolean[N];
            for (int j = 0; j < N; j++) {
                TreeNode node = all_tree_nodes.get(j);
                if (node.val == list_vals.get(i)) {
                    if (node.left != null) {
                        int left_idx = node_to_idx.get(node.left);
                        if (dp[left_idx]) {
                            new_dp[j] = true;
                        }
                    }
                    if (!new_dp[j] && node.right != null) {
                        int right_idx = node_to_idx.get(node.right);
                        if (dp[right_idx]) {
                            new_dp[j] = true;
                        }
                    }
                }
            }
            dp = new_dp;
        }

        for (boolean b : dp) {
            if (b) {
                return true;
            }
        }
        return false;
    }
}

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Problem Info

Description

Problem 1367: Linked List in Binary Tree - Detailed explanation and solution approaches.

Topics

Linked List Tree Depth-First Search Binary Tree

Companies

SoundHound

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1367

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Convert the linked list into a list of integers `list_vals` for easier access.
2
Step 2: Traverse the binary tree using BFS and store all tree nodes in a list `all_tree_nodes`.
3
Step 3: Create a hash map `node_to_idx` to map each tree node to its index in the `all_tree_nodes` list.
4
Step 4: Initialize a DP table `dp` of size N (number of tree nodes) where `dp[j]` is true if the last element of the linked list `list_vals` is found at node `all_tree_nodes[j]`.
5
Step 5: Iterate backward from the second to last element of `list_vals` to the first, updating the DP table at each step.
6
Step 6: In each inner loop iteration, check if the current tree node's value matches the current element in `list_vals`. If it does, check if either its left or right child (if they exist) has a value of `True` in the previous DP table.
7
Step 7: After the outer loop finishes, check if any element in the DP table is `True`. If so, the linked list is a subpath of the binary tree. Return `True` if a subpath exists, `False` otherwise.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is dominated by the tree traversal (BFS) and the dynamic programming calculations, which effectively visit each node in the tree at most once for each element in the linked list (which is capped by N). The space complexity comes from storing all tree nodes, the node-to-index map, and the DP table, all of which are proportional to the number of nodes in the tree.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive DFS

A recursive depth-first search (DFS) approach can be used. For each node in the tree, check if the linked list is a subpath starting at that node. If not, recursively check the left and right subtrees.

Time: O(N * min(L, H)), where N is the number of nodes in the tree, L is the length of the linked list, and H is the height of the tree. In the worst case, the time complexity is close to O(N*L)Space: O(H), where H is the height of the tree, due to the recursion stack.

Trade-offs: The recursive DFS approach is easier to implement, but can be less efficient for unbalanced trees or long linked lists. The provided DP approach is generally more efficient due to pre-processing and avoiding redundant traversals.

Iterative DFS with Stack

Implement DFS iteratively using a stack to avoid recursion overhead. Similar logic to recursive DFS but using a stack for traversal.

Time: O(N * min(L, H)), similar to recursive DFS.Space: O(N) in worst case of highly unbalanced tree (similar space usage for the stack as N number of tree nodes)

Trade-offs: Avoids recursion overhead but may still be less efficient than DP for large trees.

✨

Key Insights

1Insight 1: We can traverse the binary tree and for each node, check if the linked list is a subpath starting at that node.
2Insight 2: A dynamic programming approach can be used to efficiently determine if a subpath exists. By working backwards from the end of the linked list, we can build up a table indicating whether a portion of the linked list exists starting at a given node in the tree.
3Insight 3: Instead of recursively calling the subpath check function on each node, a more efficient approach pre-processes the tree into a list of all nodes and creates a map from node to index to avoid redundant traversals when using DP.

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None). The code handles this by returning `False` at the beginning of the `isSubPath` function.
Edge case 2: Empty linked list (head is None/empty list). This would cause L to be 0, and the backward loops would not execute properly. However, based on the constraint, the length of linked list is at least 1, therefore, we don't need to account for this.
Edge case 3: Linked list is longer than any path in the tree. The algorithm correctly handles this by iteratively checking if the next value in the linked list exists along a path in the tree.
Edge case 4: The root node's value matches the first value in the linked list. The algorithm considers this case as a potential starting point for a matching subpath.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the tree is empty.

✗Common mistake 2: Assuming that the linked list must start from the root of the tree. The linked list can be a subpath starting from any node.

✗Common mistake 3: Incorrectly implementing the tree traversal (BFS/DFS), which can lead to incorrect node indexing.

✗Common mistake 4: Not handling edge cases such as short linked lists or unbalanced trees effectively when implementing alternative recursive/iterative methods.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Searching for specific patterns or sequences in hierarchical data structures, such as XML or JSON documents. You might want to check if a certain sequence of tags appears in a particular order within the document structure.

🏦Real-world scenario 2: Network routing. Determining if a specific path (sequence of nodes) exists in a network topology. A linked list could represent a desired path, and the tree could represent the possible connections in the network.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Linked List in Binary Tree"?

For this medium-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?