What is the time complexity of Maximum Difference Between Node and Ancestor?

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

What data structures are used to solve Maximum Difference Between Node and Ancestor?

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

Which companies ask LeetCode 1026. Maximum Difference Between Node and Ancestor in interviews?

This problem is frequently asked at EPAM Systems, josh technology.

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Maximum Difference Between Node and Ancestor - LeetCode 1026 Solution

Maximum Difference Between Node and Ancestor - Complete Solution Guide

Maximum Difference Between Node and Ancestor is LeetCode problem 1026, 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 the root of a binary tree, find the maximum value v for which there exist different nodes a and b where v = |a.val - b.val| and a is an ancestor of b . A node a is an ancestor of b if either: any child of a is equal to b or any child of a is an ancestor of b . Example 1: Input: root = [8,3,10,1,6,null,14,null,null,4,7,13] Output: 7 Explanation: We have various ancestor-node differences, some of which are given below : |8 - 3| = 5 |3 - 7| = 4 |8 - 1| = 7 |10 - 13| = 3 Among all possible dif

Detailed Explanation

The problem asks us to find the maximum absolute difference between any ancestor node and its descendant node in a given binary tree. In other words, for every pair of nodes (a, b) in the tree, where 'a' is an ancestor of 'b', we calculate the absolute difference |a.val - b.val|. The goal is to find the largest such difference among all possible ancestor-descendant pairs. The input is the root node of the binary tree. The output is a single integer representing the maximum ancestor-descendant difference. The constraints limit the number of nodes to between 2 and 5000, and each node's value is between 0 and 10^5.

Solution Approach

The provided solution employs a Depth-First Search (DFS) traversal of the binary tree. During the traversal, it maintains the maximum and minimum values encountered along the path from the root to the current node. When a leaf node is reached, the difference between the current maximum and minimum values represents the maximum possible difference between ancestors and descendants along that particular path. The algorithm recursively explores all paths in the tree and updates the overall maximum difference found so far. The initial maximum and minimum values are set to the value of the root node.

Step-by-Step Algorithm

Step 1: Define a recursive helper function `dfs(node, cur_max, cur_min)` that takes the current node, the maximum value encountered so far (cur_max), and the minimum value encountered so far (cur_min) as input.
Step 2: Base Case: If the current node is null, return the difference between `cur_max` and `cur_min`. This signifies the end of a path from the root to a leaf.
Step 3: Update `cur_max` and `cur_min` to include the value of the current node. `cur_max = max(cur_max, node.val)` and `cur_min = min(cur_min, node.val)`.
Step 4: Recursively call the `dfs` function for the left and right children of the current node, passing the updated `cur_max` and `cur_min` values. `left_diff = dfs(node.left, cur_max, cur_min)` and `right_diff = dfs(node.right, cur_max, cur_min)`.
Step 5: Return the maximum of the `left_diff` and `right_diff`. This ensures that we are propagating the largest ancestor-descendant difference found so far.
Step 6: The main function `maxAncestorDiff(root)` calls the `dfs` function starting from the root node, initializing both `cur_max` and `cur_min` to the value of the root node.

Key Insights

Insight 1: We need to traverse the tree to explore all possible ancestor-descendant relationships.
Insight 2: For each node, we need to keep track of the maximum and minimum values encountered along the path from the root to that node. This is because the maximum difference will always involve the largest or smallest ancestor value encountered so far.
Insight 3: Depth-First Search (DFS) is a natural choice for traversing the tree and maintaining ancestor information.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: EPAM Systems, josh technology.

Maximum Difference Between Node and Ancestor | LeetCode 1026

Medium

Tree Depth-First Search Binary Tree

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

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 {
    public int maxAncestorDiff(TreeNode root) {
        if (root == null) {
            return 0;
        }

        return dfs(root, root.val, root.val);
    }

    private int dfs(TreeNode node, int curMax, int curMin) {
        if (node == null) {
            return curMax - curMin;
        }

        curMax = Math.max(curMax, node.val);
        curMin = Math.min(curMin, node.val);

        int leftDiff = dfs(node.left, curMax, curMin);
        int rightDiff = dfs(node.right, curMax, curMin);

        return Math.max(leftDiff, rightDiff);
    }
}

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Problem Info

Description

Problem 1026: Maximum Difference Between Node and Ancestor - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

EPAM Systemsjosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1026

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

6 steps

1
Step 1: Define a recursive helper function `dfs(node, cur_max, cur_min)` that takes the current node, the maximum value encountered so far (cur_max), and the minimum value encountered so far (cur_min) as input.
2
Step 2: Base Case: If the current node is null, return the difference between `cur_max` and `cur_min`. This signifies the end of a path from the root to a leaf.
3
Step 3: Update `cur_max` and `cur_min` to include the value of the current node. `cur_max = max(cur_max, node.val)` and `cur_min = min(cur_min, node.val)`.
4
Step 4: Recursively call the `dfs` function for the left and right children of the current node, passing the updated `cur_max` and `cur_min` values. `left_diff = dfs(node.left, cur_max, cur_min)` and `right_diff = dfs(node.right, cur_max, cur_min)`.
5
Step 5: Return the maximum of the `left_diff` and `right_diff`. This ensures that we are propagating the largest ancestor-descendant difference found so far.
6
Step 6: The main function `maxAncestorDiff(root)` calls the `dfs` function starting from the root node, initializing both `cur_max` and `cur_min` to the value of the root 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(h)

📝

Why?

The time complexity is O(n) because each node is visited once during the DFS traversal. The space complexity is determined by the maximum depth of the recursion call stack, which is equal to the height of the tree. The function call stack grows proportionally to the height of 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

Iterative DFS

An alternative approach is to implement DFS iteratively using a stack. This can avoid the overhead of recursive function calls and might be slightly more memory-efficient in some cases.

Time: O(n)Space: O(h), where h is the height of the tree. In the worst case (skewed tree), it could be O(n).

Trade-offs: Iterative DFS can be more complex to implement than recursive DFS, but it can be beneficial for very deep trees where recursion might lead to stack overflow errors. The stack would need to maintain both node and min/max information.

✨

Key Insights

1Insight 1: We need to traverse the tree to explore all possible ancestor-descendant relationships.
2Insight 2: For each node, we need to keep track of the maximum and minimum values encountered along the path from the root to that node. This is because the maximum difference will always involve the largest or smallest ancestor value encountered so far.
3Insight 3: Depth-First Search (DFS) is a natural choice for traversing the tree and maintaining ancestor information.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The provided solutions handle this case by returning 0, which is the correct answer in this scenario.
Edge case 2: A tree with only one node. The given problem statement says there should be at least two nodes so technically this isn't possible. The solution initializes min and max with the root, which resolves this case to a diff of 0 should it happen.
Edge case 3: Tree with all nodes having the same value. The algorithm will correctly return 0, as the maximum and minimum values will always be the same.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not updating the `cur_max` and `cur_min` values correctly during the DFS traversal. This can lead to incorrect calculations of the ancestor-descendant differences.

✗Common mistake 2: Failing to handle the base case (null node) properly. This can result in incorrect results or runtime errors.

✗Common mistake 3: Assuming that the maximum difference will always occur between the root and a leaf node. This is not always the case; the maximum difference can occur between any ancestor-descendant pair.

✅

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: Analyzing organizational hierarchies: In a company's organizational structure represented as a tree, this algorithm could be used to find the maximum difference in levels of responsibility or experience between a manager (ancestor) and a subordinate (descendant). The node value could represent the seniority or salary of an employee.

🏦Real-world scenario 2: Phylogenetic tree analysis: In biology, phylogenetic trees represent the evolutionary relationships between different species. The node values could represent some measurable characteristic of the species (e.g., size, genetic diversity). This algorithm could be used to find the maximum difference in that characteristic between ancestral and descendant species.

💡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

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

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: October 5, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

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 "Maximum Difference Between Node and Ancestor"?

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

🤖 Need help with this problem?