Top View of Binary Tree in DSA Rust - Deep Dive

Overview - Top View of Binary Tree

What is it?

The top view of a binary tree is the set of nodes visible when the tree is seen from above. Imagine looking down on the tree and noting which nodes you can see without any other nodes blocking them. This view shows one node per vertical line of the tree, from leftmost to rightmost.

Why it matters

Without the concept of top view, we might miss understanding how a tree looks from different perspectives, which is important in visualization and certain algorithms. It helps in problems where we want to capture the outline or shape of the tree from above, useful in graphics, spatial data, and hierarchical data analysis.

Where it fits

Before learning top view, you should understand binary trees and tree traversals like level order. After this, you can explore related views like bottom view, left view, and right view of trees, and then move on to more complex tree problems like vertical order traversal.

Mental Model

Core Idea

The top view of a binary tree shows the first node encountered at each vertical position when looking down from above.

Think of it like...

Imagine standing on a balcony looking straight down at a garden with trees planted in rows. You only see the tops of the trees that are not hidden behind others in your line of sight.

          (0)
         Root
        /    \
   (-1)Left  Right(+1)

Vertical lines: -1    0    +1
Top view picks one node per vertical line, the topmost visible.

Build-Up - 6 Steps

1

Foundation

Understanding Binary Tree Structure

Concept: Learn what a binary tree is and how nodes connect.

A binary tree is a structure where each node has up to two children: left and right. Each node holds a value and pointers to its children. The root is the top node. Example: struct Node { value: i32, left: Option>, right: Option>, } This structure allows hierarchical data storage.

Result

You can represent any binary tree with nodes linked to left and right children.

Understanding the basic tree structure is essential before exploring views like top view, which depend on node positions.

2

Foundation

Concept of Horizontal Distance

Concept: Assign horizontal distances to nodes relative to root.

We assign a horizontal distance (HD) to each node: root has HD=0, left child HD=parent HD - 1, right child HD=parent HD + 1. This helps group nodes vertically. Example: Root (HD=0) Left child (HD=-1) Right child (HD=+1)

Result

Each node has a unique horizontal distance that helps identify its vertical line.

Horizontal distance is the key to grouping nodes for top view calculation.

3

Intermediate

Level Order Traversal with Horizontal Distance

🤔Before reading on: Do you think a simple preorder traversal can find the top view? Commit to yes or no.

Concept: Use level order traversal to visit nodes by depth and track horizontal distances.

We use a queue to traverse nodes level by level. For each node, we track its HD. This ensures we see nodes from top to bottom at each vertical line. Pseudocode: - Start with root at HD=0 - Enqueue (node, HD) - While queue not empty: - Dequeue (node, HD) - If HD not seen before, record node - Enqueue left child with HD-1 - Enqueue right child with HD+1

Result

We get the first node encountered at each horizontal distance, which forms the top view.

Level order traversal combined with HD tracking ensures the topmost nodes per vertical line are captured.

4

Intermediate

Storing and Printing Top View Nodes

🤔Before reading on: Should we store nodes in a map keyed by horizontal distance or by node value? Commit to your answer.

Concept: Use a map/dictionary keyed by horizontal distance to store the first node encountered.

We keep a map: HD -> node value. When we visit a node, if HD is not in map, we add it. After traversal, we sort keys and print values. Example: HD: -2 -> 4 HD: -1 -> 2 HD: 0 -> 1 HD: 1 -> 3 HD: 2 -> 7 Print in order: 4 2 1 3 7

Result

The top view nodes printed from leftmost to rightmost horizontal distance.

Mapping HD to first node ensures we only keep the topmost node per vertical line.

5

Advanced

Rust Implementation of Top View Algorithm

🤔Before reading on: Do you think Rust's ownership model complicates storing nodes in a queue? Commit to yes or no.

Concept: Implement the top view algorithm in Rust using safe ownership and borrowing.

We define a Node struct with Box pointers. Use VecDeque for queue. Store HD and node references. Use a BTreeMap for ordered HD keys. Code snippet: use std::collections::{VecDeque, BTreeMap}; struct Node { value: i32, left: Option>, right: Option>, } fn top_view(root: &Option>) -> Vec { let mut queue = VecDeque::new(); let mut map = BTreeMap::new(); if let Some(node) = root { queue.push_back((node.as_ref(), 0)); } while let Some((node, hd)) = queue.pop_front() { map.entry(hd).or_insert(node.value); if let Some(left) = &node.left { queue.push_back((left.as_ref(), hd - 1)); } if let Some(right) = &node.right { queue.push_back((right.as_ref(), hd + 1)); } } map.values().cloned().collect() }

Result

The function returns a vector of node values representing the top view from left to right.

Rust's ownership rules require careful use of references and collections, but the algorithm remains clear and efficient.

6

Expert

Handling Large Trees and Edge Cases

🤔Before reading on: Do you think nodes with the same horizontal distance but different depths can both appear in top view? Commit to yes or no.

Concept: Understand that only the topmost node per horizontal distance appears, and optimize for large trees.

If multiple nodes share the same HD, only the first encountered at the smallest depth is in top view. For large trees, use efficient data structures like BTreeMap for ordered keys and VecDeque for BFS queue. Also, handle empty trees and single-node trees gracefully. Consider memory and runtime: BFS ensures O(n) time, and BTreeMap keeps keys sorted for output.

Result

Robust top view calculation that works efficiently even for large or skewed trees.

Knowing how depth and horizontal distance interact prevents incorrect inclusion of lower nodes in top view.

Under the Hood

The algorithm uses breadth-first search (BFS) to visit nodes level by level, tracking horizontal distances (HD) from the root. A map stores the first node encountered at each HD, ensuring only the topmost node per vertical line is recorded. BFS guarantees nodes are visited in order of increasing depth, so the first node at each HD is the top view node.

Why designed this way?

BFS was chosen because preorder or inorder traversals do not guarantee visiting nodes in top-to-bottom order per vertical line. Tracking HD allows grouping nodes by vertical lines. Using a map keyed by HD ensures quick lookup and insertion. This design balances clarity, correctness, and efficiency.

┌───────────────┐
│    Root (HD=0)│
└──────┬────────┘
       │
  ┌────┴────┐
  │         │
Left(HD=-1) Right(HD=+1)

Queue stores (node, HD) pairs
Map stores HD -> first node value
BFS visits nodes level-wise
First node per HD recorded
Output sorted by HD keys

Myth Busters - 3 Common Misconceptions

Quick: Does the top view include all nodes at a horizontal distance or only the first encountered? Commit to your answer.

Common Belief: Top view includes all nodes that share the same horizontal distance.

Reality: Top view includes only the first node encountered at each horizontal distance during level order traversal.

Why it matters: Including all nodes at the same horizontal distance would produce an incorrect view, showing nodes hidden behind others.

Quick: Can a simple inorder traversal find the top view correctly? Commit yes or no.

Common Belief: Inorder traversal can be used to find the top view of a binary tree.

Reality: Inorder traversal does not guarantee visiting nodes in top-to-bottom order per vertical line, so it cannot reliably find the top view.

Why it matters: Using inorder traversal leads to incorrect or incomplete top view results.

Quick: Is horizontal distance the same as depth? Commit yes or no.

Common Belief: Horizontal distance and depth are the same or interchangeable concepts.

Reality: Horizontal distance measures left-right position relative to root; depth measures vertical level from root. They are different and both are needed for top view.

Why it matters: Confusing these leads to wrong grouping of nodes and incorrect top view.

Expert Zone

1

The choice of data structure for storing horizontal distances affects performance; BTreeMap keeps keys sorted automatically, avoiding extra sorting steps.

2

In trees with overlapping nodes at the same HD and depth, the node encountered first in BFS order is chosen, which depends on enqueue order of children.

3

Rust's ownership and borrowing rules require careful handling of node references in queues to avoid cloning or unsafe code.

When NOT to use

Top view is not suitable when you need to see all nodes at each vertical line (use vertical order traversal instead). For bottom view, which shows nodes visible from below, a different approach is needed.

Production Patterns

Top view is used in graphical tree visualizations, spatial indexing, and in problems requiring skyline or silhouette extraction from hierarchical data. It is often combined with other views for comprehensive tree analysis.

Connections

Vertical Order Traversal

Builds on the idea of horizontal distance grouping but includes all nodes per vertical line sorted by depth.

Understanding top view helps grasp vertical order traversal, which generalizes the concept by including all nodes, not just the topmost.

Breadth-First Search (BFS)

Top view uses BFS to visit nodes level by level ensuring correct order of node discovery.

Knowing BFS is crucial because it guarantees visiting nodes in increasing depth order, which is essential for correct top view calculation.

Urban Skyline Problem (Computational Geometry)

Both problems involve finding visible outlines from a viewpoint, one in trees and one in city skylines.

Recognizing the similarity between top view and skyline problems reveals how visibility and occlusion concepts apply across computer science and geometry.

Common Pitfalls

#1 Using preorder or inorder traversal to find top view.

Wrong approach:

fn top_view_preorder(node: &Option>, hd: i32, map: &mut HashMap) {
if let Some(n) = node {
map.entry(hd).or_insert(n.value);
top_view_preorder(&n.left, hd - 1, map);
top_view_preorder(&n.right, hd + 1, map);
}
}

Correct approach:

Use BFS with queue to ensure level order traversal:

fn top_view(root: &Option>) -> Vec {
let mut queue = VecDeque::new();
let mut map = BTreeMap::new();
if let Some(node) = root {
queue.push_back((node.as_ref(), 0));
}
while let Some((node, hd)) = queue.pop_front() {
map.entry(hd).or_insert(node.value);
if let Some(left) = &node.left {
queue.push_back((left.as_ref(), hd - 1));
}
if let Some(right) = &node.right {
queue.push_back((right.as_ref(), hd + 1));
}
}
map.values().cloned().collect()
}

Root cause:

Preorder or inorder traversals do not visit nodes in top-to-bottom order per vertical line, so they cannot correctly identify the topmost nodes.

#2 Not sorting horizontal distances before printing top view.

Wrong approach:

for (hd, val) in map.iter() {
println!("{}", val);
}

Correct approach:

Use a BTreeMap or sort keys before printing:

for hd in map.keys().sorted() {
println!("{}", map[hd]);
}

Root cause:

HashMaps do not guarantee order, so printing without sorting leads to unordered output.

#3 Confusing horizontal distance with depth and using depth as key.

Wrong approach:

map.entry(depth).or_insert(node.value);

Correct approach:

map.entry(horizontal_distance).or_insert(node.value);

Root cause:

Mixing depth and horizontal distance causes incorrect grouping of nodes, breaking the top view logic.

Key Takeaways

Top view of a binary tree shows the first node visible at each vertical line when viewed from above.

Horizontal distance from the root is used to group nodes vertically, while level order traversal ensures nodes are visited top-down.

Using a map keyed by horizontal distance and BFS traversal guarantees correct and efficient top view extraction.

Rust's ownership model requires careful handling of node references but does not complicate the core algorithm.

Understanding top view lays the foundation for related tree views and spatial visibility problems.

Practice

1. What does the top view of a binary tree represent?

easy

A. Nodes visible when the tree is viewed from the top

B. Nodes visible when the tree is viewed from the left side

C. Nodes visible when the tree is viewed from the bottom

D. Nodes visible when the tree is viewed from the right side

Solution

1. Step 1: Understand the top view concept

   The top view of a binary tree shows all nodes that are visible when looking down vertically from above the tree.

2. Step 2: Compare with other views

   Left, right, and bottom views show nodes visible from those sides, not from the top.

3. Final Answer:

   Nodes visible when the tree is viewed from the top -> Option A

4. Quick Check:

   Top view = nodes seen from above [OK]

Hint: Top view shows nodes visible from above the tree [OK]

Common Mistakes:

• Confusing top view with left or right view
• Thinking bottom view is the same as top view

2. Which Rust data structure is best suited to store nodes with their horizontal distances for computing the top view of a binary tree?

easy

A. HashMap

B. Vec<Node>

C. LinkedList<Node>

D. BinaryHeap<Node>

Solution

1. Step 1: Identify the need for horizontal distance mapping

   Top view requires tracking nodes by their horizontal distance from the root, which can be negative or positive integers.

2. Step 2: Choose a data structure for key-value mapping

   A HashMap with key as horizontal distance (i32) and value as Node is ideal for quick lookups and updates.

3. Final Answer:

   HashMap -> Option A

4. Quick Check:

   Use HashMap for horizontal distance mapping [OK]

Hint: Use HashMap to map horizontal distance to nodes [OK]

Common Mistakes:

• Using Vec which doesn't map keys to values
• Using LinkedList which is inefficient for lookups
• Using BinaryHeap which is for priority queues

3. Given the following Rust code snippet for top view traversal, what will be printed?

struct Node {
    val: i32,
    left: Option,
    right: Option,
}

// Tree:
//       1
//      / \
//     2   3
//      \   \
//       4   5

fn main() {
    let root = Node {
        val: 1,
        left: Some(Box::new(Node {
            val: 2,
            left: None,
            right: Some(Box::new(Node { val: 4, left: None, right: None })),
        })),
        right: Some(Box::new(Node {
            val: 3,
            left: None,
            right: Some(Box::new(Node { val: 5, left: None, right: None })),
        })),
    };
    // Assume top_view function prints top view nodes in order from left to right
    top_view(&root);
}

medium

A. [2, 4, 1, 3, 5]

B. [2, 1, 3, 5]

C. [4, 2, 1, 3, 5]

D. [1, 2, 3]

Solution

1. Step 1: Assign horizontal distances to nodes

   Root (1) is at 0, left child (2) at -1, right child (3) at +1, node 4 at 0 (right child of 2), node 5 at +2 (right child of 3).

2. Step 2: Determine top view nodes by horizontal distance

   For each horizontal distance, the first node encountered from top is chosen: -1: 2, 0: 1, +1: 3, +2: 5. Node 4 at 0 is below 1, so ignored.

3. Final Answer:

   [2, 1, 3, 5] -> Option B

4. Quick Check:

   Top view nodes by horizontal distance = [2, 1, 3, 5] [OK]

Hint: Pick first node at each horizontal distance from top [OK]

Common Mistakes:

• Including nodes below first at same horizontal distance
• Ignoring horizontal distances for nodes
• Confusing order of output

4. The following Rust code attempts to compute the top view of a binary tree but has a bug. What is the error?

fn top_view(root: &Node) {
    let mut map = std::collections::HashMap::new();
    let mut queue = std::collections::VecDeque::new();
    queue.push_back((root, 0));

    while let Some((node, hd)) = queue.pop_front() {
        // Insert node if horizontal distance not present
        map.entry(hd).or_insert(node.val);

        if let Some(left) = &node.left {
            queue.push_back((left, hd - 1));
        }
        if let Some(right) = &node.right {
            queue.push_back((right, hd + 1));
        }
    }

    for key in map.keys() {
        println!("{}", map[key]);
    }
}

medium

A. The code inserts duplicate nodes for same horizontal distance

B. The queue is not initialized properly

C. The map keys are not sorted before printing, so output order is random

D. The code does not handle empty tree (root is None)

Solution

1. Step 1: Analyze map insertion and printing

   The code inserts nodes correctly using or_insert to keep first node per horizontal distance.

2. Step 2: Check output order

   HashMap keys are unordered, so iterating with map.keys() prints nodes in random order, not left to right.

3. Final Answer:

   The map keys are not sorted before printing, so output order is random -> Option C

4. Quick Check:

   Sort keys before printing top view [OK]

Hint: Sort horizontal distances before printing top view [OK]

Common Mistakes:

• Assuming HashMap keys are ordered
• Ignoring empty tree case (not asked here)
• Thinking queue initialization is wrong

5. You want to find the top view of a binary tree but only have the tree nodes and no horizontal distance stored. Which approach correctly computes the top view in Rust? A) Use a recursive function passing horizontal distance and level, store first node per horizontal distance with smallest level. B) Use a stack to traverse nodes and store last node per horizontal distance. C) Traverse the tree in preorder and print nodes as you visit. D) Use a queue but ignore horizontal distances and print nodes in insertion order.

hard

A. Preorder traversal printing nodes directly

B. Use stack and store last node per horizontal distance

C. Use queue ignoring horizontal distances

D. Use recursion with horizontal distance and level to track top nodes

Solution

1. Step 1: Understand top view requires horizontal distance and level

   Top view needs to know horizontal distance and the vertical level to pick the topmost node at each horizontal distance.

2. Step 2: Choose approach that tracks horizontal distance and level

   Recursive traversal passing horizontal distance and level, storing first node per horizontal distance with smallest level, correctly computes top view.

3. Final Answer:

   Use recursion with horizontal distance and level to track top nodes -> Option D

4. Quick Check:

   Track horizontal distance and level recursively [OK]

Hint: Track horizontal distance and level to find top nodes [OK]

Common Mistakes:

• Ignoring horizontal distance in traversal
• Using stack without tracking levels
• Printing nodes in preorder without filtering