What is the time complexity of Vertical Order Traversal of a Binary Tree?

The optimal solution has a time complexity of O(W*H*log(H)).

What data structures are used to solve Vertical Order Traversal of a Binary Tree?

This problem uses Hash Table, Tree, Depth-First Search, Breadth-First Search, Sorting, Binary Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Vertical Order Traversal of a Binary Tree?

Insight 1: We can use a Breadth-First Search (BFS) to traverse the tree level by level, assigning coordinates to each node as we visit them. Insight 2: A hash map (or map in C++) is essential to store nodes based on their column number. This allows us to easily group nodes in the same column.

Which companies ask LeetCode 987. Vertical Order Traversal of a Binary Tree in interviews?

This problem is frequently asked at Deliveroo, DoorDash, Samsung, eBay.

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Vertical Order Traversal of a Binary Tree - LeetCode 987 Solution

Vertical Order Traversal of a Binary Tree - Complete Solution Guide

Vertical Order Traversal of a Binary Tree is LeetCode problem 987, a Hard 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, calculate the vertical order traversal of the binary tree. For each node at position (row, col) , its left and right children will be at positions (row + 1, col - 1) and (row + 1, col + 1) respectively. The root of the tree is at (0, 0) . The vertical order traversal of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same

Detailed Explanation

The problem asks us to perform a vertical order traversal of a binary tree. Each node in the tree has a coordinate (row, col), with the root at (0, 0). A node's left child is at (row+1, col-1), and its right child is at (row+1, col+1). We need to return a list of lists, where each inner list contains the nodes in a particular column, sorted from top to bottom. If nodes are in the same row and column, they should be sorted by their values.

Solution Approach

The solution uses a BFS traversal to visit each node in the tree, keeping track of the row and column of each node. A hash map (column_table) is used to store a list of (row, value) pairs for each column. After traversing the entire tree, the keys of the hash map (representing the columns) are sorted. For each column, the (row, value) pairs are sorted based on row first and then by value. Finally, the sorted values are extracted and added to the result list.

Step-by-Step Algorithm

Step 1: Initialize a queue for BFS traversal and a hash map (column_table) to store nodes based on column number.
Step 2: Enqueue the root node with coordinates (0, 0).
Step 3: While the queue is not empty:
Step 4: Dequeue a node (node, row, col) from the queue.
Step 5: Store the (row, node.val) in the column_table at key 'col'.
Step 6: Enqueue the left child (if it exists) with coordinates (row+1, col-1).
Step 7: Enqueue the right child (if it exists) with coordinates (row+1, col+1).
Step 8: After the BFS is complete, sort the keys (column numbers) of the column_table.
Step 9: Iterate through the sorted column numbers:
Step 10: Sort the (row, value) pairs in the current column based on row (ascending) and value (ascending).
Step 11: Extract the values from the sorted pairs and add them to a list representing the column's vertical order.
Step 12: Add the list of values for the current column to the result list.
Step 13: Return the result list.

Key Insights

Insight 1: We can use a Breadth-First Search (BFS) to traverse the tree level by level, assigning coordinates to each node as we visit them.
Insight 2: A hash map (or map in C++) is essential to store nodes based on their column number. This allows us to easily group nodes in the same column.
Insight 3: We need to sort the nodes within each column based on their row and value. This requires using a custom sorting function or lambda expression.

Complexity Analysis

Time Complexity: O(W*H*log(H))

Space Complexity: O(N)

Topics

This problem involves: Hash Table, Tree, Depth-First Search, Breadth-First Search, Sorting, Binary Tree.

Companies

Asked at: Deliveroo, DoorDash, Samsung, eBay.

Vertical Order Traversal of a Binary Tree | LeetCode 987

Hard

Hash Table Tree Depth-First Search Breadth-First Search Sorting Binary Tree

Time: O(W*H*log(H))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;
 *     }
 * }
 */
import java.util.*;

class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }

        Map<Integer, List<Pair<Integer, Integer>>> columnTable = new HashMap<>();
        Queue<Triple<TreeNode, Integer, Integer>> queue = new LinkedList<>();
        queue.offer(new Triple<>(root, 0, 0));

        while (!queue.isEmpty()) {
            Triple<TreeNode, Integer, Integer> triple = queue.poll();
            TreeNode node = triple.node;
            int row = triple.row;
            int col = triple.col;

            if (!columnTable.containsKey(col)) {
                columnTable.put(col, new ArrayList<>());
            }
            columnTable.get(col).add(new Pair<>(row, node.val));

            if (node.left != null) {
                queue.offer(new Triple<>(node.left, row + 1, col - 1));
            }

            if (node.right != null) {
                queue.offer(new Triple<>(node.right, row + 1, col + 1));
            }
        }

        List<List<Integer>> result = new ArrayList<>();
        List<Integer> sortedKeys = new ArrayList<>(columnTable.keySet());
        Collections.sort(sortedKeys);

        for (int key : sortedKeys) {
            List<Pair<Integer, Integer>> columnValues = columnTable.get(key);
            Collections.sort(columnValues, (a, b) -> {
                if (a.getKey() != b.getKey()) {
                    return a.getKey() - b.getKey();
                } else {
                    return a.getValue() - b.getValue();
                }
            });

            List<Integer> columnResult = new ArrayList<>();
            for (Pair<Integer, Integer> pair : columnValues) {
                columnResult.add(pair.getValue());
            }
            result.add(columnResult);
        }

        return result;
    }
}

class Triple<TreeNode, Integer, Integer> {
    TreeNode node;
    int row;
    int col;

    public Triple(TreeNode node, int row, int col) {
        this.node = node;
        this.row = row;
        this.col = col;
    }
}

class Pair<Integer, Integer> {
    Integer key;
    Integer value;

    public Pair(Integer key, Integer value) {
        this.key = key;
        this.value = value;
    }

    public Integer getKey() {
        return key;
    }

    public Integer getValue() {
        return value;
    }
}

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Problem Info

Description

Problem 987: Vertical Order Traversal of a Binary Tree - Detailed explanation and solution approaches.

Topics

Hash Table Tree Depth-First Search Breadth-First Search Sorting Binary Tree

Companies

DeliverooDoorDashSamsungeBay

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:987

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

13 steps

1
Step 1: Initialize a queue for BFS traversal and a hash map (column_table) to store nodes based on column number.
2
Step 2: Enqueue the root node with coordinates (0, 0).
3
Step 3: While the queue is not empty:
4
Step 4: Dequeue a node (node, row, col) from the queue.
5
Step 5: Store the (row, node.val) in the column_table at key 'col'.
6
Step 6: Enqueue the left child (if it exists) with coordinates (row+1, col-1).
7
Step 7: Enqueue the right child (if it exists) with coordinates (row+1, col+1).
8
Step 8: After the BFS is complete, sort the keys (column numbers) of the column_table.
9
Step 9: Iterate through the sorted column numbers:
10
Step 10: Sort the (row, value) pairs in the current column based on row (ascending) and value (ascending).
11
Step 11: Extract the values from the sorted pairs and add them to a list representing the column's vertical order.
12
Step 12: Add the list of values for the current column to the result list.
13
Step 13: Return the result list.

💡

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(W*H*log(H))

💾

Space Complexity

O(N)

📝

Why?

The BFS traversal visits each node once. The space complexity is determined by the maximum number of nodes stored in the queue during the BFS and in the hash map. Sorting each column, at worst, takes O(N log N) time. Hence, overall time complexity is O(N log N).

📖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

Depth-First Search (DFS)

Instead of BFS, we could use DFS to traverse the tree and calculate the column and row for each node. The rest of the process (sorting the columns and values within each column) would remain the same.

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

Trade-offs: DFS might be slightly less intuitive for this problem compared to BFS because it's inherently not level-order. The time and space complexity would remain largely the same.

✨

Key Insights

1Insight 1: We can use a Breadth-First Search (BFS) to traverse the tree level by level, assigning coordinates to each node as we visit them.
2Insight 2: A hash map (or map in C++) is essential to store nodes based on their column number. This allows us to easily group nodes in the same column.
3Insight 3: We need to sort the nodes within each column based on their row and value. This requires using a custom sorting function or lambda expression.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Tree

Hierarchical structure with nodes. Like a family tree!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty tree. The code correctly handles this case by returning an empty list.
Edge case 2: Tree with only one node. The code correctly places the node in column 0 at row 0.
Edge case 3: Skewed tree (all nodes on one side). The code correctly handles this by assigning appropriate column numbers and sorting the nodes within each column.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the nodes within each column by both row and value. The problem statement specifically requires sorting by value if nodes are at the same row and column.

✗Common mistake 2: Not properly calculating the column number. The column number should decrease for left children and increase for right children.

✗Common mistake 3: Forgetting to sort the columns before combining the results. The final result should be ordered by column number from left to right.

✅

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: Displaying a hierarchical structure in a vertical layout, like an organizational chart or a file system tree view.

🏦Real-world scenario 2: Visualizing data from a database with hierarchical relationships. The vertical order traversal could determine the order in which elements are displayed.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: August 27, 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 do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Vertical Order Traversal of a Binary Tree"?

For this hard-level Hash Table 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?