What is the time complexity of Longest ZigZag Path in a Binary Tree?

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

What data structures are used to solve Longest ZigZag Path in a Binary Tree?

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

What are the key insights for solving Longest ZigZag Path in a Binary Tree?

Insight 1: Dynamic Programming: The problem can be solved efficiently using dynamic programming by calculating the longest ZigZag path starting at each node in both left and right directions. This avoids redundant calculations. Insight 2: Post-order Traversal: A post-order traversal of the tree is crucial because to determine the length of the ZigZag path originating from a node, we need the lengths of ZigZag paths starting at its children.

Which companies ask LeetCode 1372. Longest ZigZag Path in a Binary Tree in interviews?

This problem is frequently asked at eBay.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Longest ZigZag Path in a Binary Tree - LeetCode 1372 Solution

Longest ZigZag Path in a Binary Tree - Complete Solution Guide

Longest ZigZag Path in a Binary Tree is LeetCode problem 1372, 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

You are given the root of a binary tree. A ZigZag path for a binary tree is defined as follow: Choose any node in the binary tree and a direction (right or left). If the current direction is right, move to the right child of the current node; otherwise, move to the left child. Change the direction from right to left or from left to right. Repeat the second and third steps until you can't move in the tree. Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of

Detailed Explanation

The problem asks us to find the longest ZigZag path in a given binary tree. A ZigZag path is defined as a path starting from any node and direction (left or right), and alternating directions (left to right or right to left) until we cannot move further. The length of the path is the number of nodes visited minus 1. The task is to find the maximum length of such a ZigZag path within the given tree.

Solution Approach

The provided solution uses a recursive Depth-First Search (DFS) approach with dynamic programming. The `dfs` function (or `solve` function in Java/C/C++) computes the length of the longest ZigZag paths starting at a given node, going both left and right. It stores the results (the left and right path lengths) and also keeps track of the overall maximum length found so far using a global variable. The post-order traversal ensures that the children are processed before their parents, allowing us to build the solution from the bottom up.

Step-by-Step Algorithm

Step 1: Define a recursive function (dfs/solve) that takes a node as input and returns a pair of integers: (left_path_length, right_path_length). These represent the longest zigzag path lengths starting at the given node going left and right respectively.
Step 2: Base Case: If the node is null, return (-1, -1). This ensures that when we add 1 to the length, it becomes 0 for the starting node (as the initial edge doesn't exist).
Step 3: Recursively call the function for the left and right children of the current node.
Step 4: Calculate the length of the ZigZag path starting at the current node and going left. This is equal to 1 + the length of the longest ZigZag path starting at the left child and going right (left_path = 1 + right_path_from_left_child).
Step 5: Calculate the length of the ZigZag path starting at the current node and going right. This is equal to 1 + the length of the longest ZigZag path starting at the right child and going left (right_path = 1 + left_path_from_right_child).
Step 6: Update the global variable `max_length` with the maximum of its current value, `left_path`, and `right_path`.
Step 7: Return the pair (left_path, right_path) for the current node.

Key Insights

Insight 1: Dynamic Programming: The problem can be solved efficiently using dynamic programming by calculating the longest ZigZag path starting at each node in both left and right directions. This avoids redundant calculations.
Insight 2: Post-order Traversal: A post-order traversal of the tree is crucial because to determine the length of the ZigZag path originating from a node, we need the lengths of ZigZag paths starting at its children.
Insight 3: Recursive Approach: The problem lends itself naturally to a recursive implementation, where each recursive call computes the ZigZag path lengths starting from a given node.
Insight 4: State at each Node: For each node, we need to maintain two states: the longest zigzag path starting from the node going left and the longest zigzag path starting from the node going right. This information allows us to build the solution recursively.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Tree, Depth-First Search, Binary Tree.

Companies

Asked at: eBay.

Longest ZigZag Path in a Binary Tree | LeetCode 1372

Medium

Dynamic Programming Tree Depth-First Search Binary Tree

Time: O(n)Space: O(n)

Solution Code

Available in:

/**
 * 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 {
    private int max_length = 0;

    public int longestZigZag(TreeNode root) {
        max_length = 0;
        solve(root);
        return max_length;
    }

    private int[] solve(TreeNode node) {
        /**
         * Returns a tuple (left_path, right_path) for the given node.
         * left_path: length of the longest zigzag path starting at this node and going left.
         * right_path: length of the longest zigzag path starting at this node and going right.
         */
        if (node == null) {
            return new int[]{-1, -1};
        }

        // Post-order traversal: process children first
        int[] left_child_paths = solve(node.left);
        int left_from_left_child = left_child_paths[0];
        int right_from_left_child = left_child_paths[1];

        int[] right_child_paths = solve(node.right);
        int left_from_right_child = right_child_paths[0];
        int right_from_right_child = right_child_paths[1];

        // A zigzag path starting at the current node and going left
        // is 1 (for the edge to the left child) plus the length of the
        // path that starts at the left child and goes right.
        int left_path = 1 + right_from_left_child;

        // Similarly, for the path going right.
        int right_path = 1 + left_from_right_child;

        // The overall maximum could be any of these paths starting at any node.
        max_length = Math.max(max_length, Math.max(left_path, right_path));

        return new int[]{left_path, right_path};
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1372: Longest ZigZag Path in a Binary Tree - Detailed explanation and solution approaches.

Topics

Dynamic Programming Tree Depth-First Search Binary Tree

Companies

eBay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1372

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: Define a recursive function (dfs/solve) that takes a node as input and returns a pair of integers: (left_path_length, right_path_length). These represent the longest zigzag path lengths starting at the given node going left and right respectively.
2
Step 2: Base Case: If the node is null, return (-1, -1). This ensures that when we add 1 to the length, it becomes 0 for the starting node (as the initial edge doesn't exist).
3
Step 3: Recursively call the function for the left and right children of the current node.
4
Step 4: Calculate the length of the ZigZag path starting at the current node and going left. This is equal to 1 + the length of the longest ZigZag path starting at the left child and going right (left_path = 1 + right_path_from_left_child).
5
Step 5: Calculate the length of the ZigZag path starting at the current node and going right. This is equal to 1 + the length of the longest ZigZag path starting at the right child and going left (right_path = 1 + left_path_from_right_child).
6
Step 6: Update the global variable `max_length` with the maximum of its current value, `left_path`, and `right_path`.
7
Step 7: Return the pair (left_path, right_path) for the current node.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Time Complexity: The time complexity is O(n) because we visit each node in the tree exactly once during the Depth-First Search (DFS) traversal. Space Complexity: The space complexity is O(n) in the worst case, where n is the number of nodes in the tree. This is due to the recursion stack. In a skewed tree, the recursion depth can reach n. In a balanced tree, the recursion depth will be O(log n), but the global max_length variable remains constant.

📖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

Iterative DFS

An iterative DFS approach using a stack can be used to avoid recursion. This might reduce the stack space usage but increases the implementation complexity.

Time: O(n)Space: O(n)

Trade-offs: Iterative DFS avoids the risk of stack overflow errors that can occur with deep recursion, but it generally requires more code and can be harder to read and debug.

✨

Key Insights

1Insight 1: Dynamic Programming: The problem can be solved efficiently using dynamic programming by calculating the longest ZigZag path starting at each node in both left and right directions. This avoids redundant calculations.
2Insight 2: Post-order Traversal: A post-order traversal of the tree is crucial because to determine the length of the ZigZag path originating from a node, we need the lengths of ZigZag paths starting at its children.
3Insight 3: Recursive Approach: The problem lends itself naturally to a recursive implementation, where each recursive call computes the ZigZag path lengths starting from a given node.
4Insight 4: State at each Node: For each node, we need to maintain two states: the longest zigzag path starting from the node going left and the longest zigzag path starting from the node going right. This information allows us to build the solution recursively.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty Tree: If the input tree is empty (root is null), the algorithm correctly returns 0 because no path exists. The base case handles this scenario gracefully.
Edge case 2: Single Node Tree: If the tree consists of only one node, the algorithm returns 0 because there are no edges to form a ZigZag path. The initial values are set to -1 which upon adding 1 become 0, indicating a path length of 0.
Edge case 3: Skewed Tree: A skewed tree (where all nodes are only left children or right children) can still be processed correctly. The algorithm effectively explores the paths and keeps track of the maximum length.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect base case: Failing to properly handle the base case (null node) can lead to incorrect path lengths and ultimately a wrong result. Returning -1 in the base case is important for the calculation to work correctly.

✗Common mistake 2: Not updating the global maximum length: Forgetting to update the global `max_length` variable within the recursive function will result in the algorithm not finding the overall maximum ZigZag path in the entire tree. Only returning the path lengths to the parent is not sufficient.

✗Common mistake 3: Incorrectly calculating path lengths: Adding 1 in the wrong place, or not taking into account whether a left or right turn should be performed leads to the incorrect result.

✗Common mistake 4: Not understanding post-order traversal: Failing to recognize that children should be processed first leads to issues when calculating the path lengths at a node, since child node information is needed.

✅

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: Network Routing: Finding the longest zigzag path can be conceptually related to finding efficient routing paths in network graphs, where the direction of the path might represent different network protocols or path attributes. In this analogy, the binary tree can be seen as representing the network connections.

🏦Real-world scenario 2: Decision Trees: In machine learning, decision trees can have 'zigzag' paths depending on the decisions made at each node. Finding the longest such path, if certain constraints are applied to 'zig' and 'zag' decisions, could represent the most complex decision-making process within the model.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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 21, 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

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Longest ZigZag Path in a Binary Tree"?

For this medium-level Dynamic Programming 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?