Why Top View of Binary Tree in DSA Rust?

The Big Idea

Discover how to see the hidden shape of a tree from above without climbing it!

The Scenario

Imagine you have a tall tree in your backyard. You want to see which branches you can spot if you look at the tree from the top, but you only have a ladder to climb and look from the side.

Trying to figure out the top view by climbing and guessing which branches are visible is like manually checking each branch from different angles.

The Problem

Manually checking each branch from the side is slow and confusing. You might miss some branches hidden behind others or count the same branch multiple times.

It's hard to keep track of which branches are visible from the top without a clear plan.

The Solution

The top view of a binary tree shows exactly which nodes (branches) are visible when you look at the tree from above.

Using a smart method, we can find these visible nodes by tracking their horizontal positions and choosing the first node we see at each position.

This way, we get a clear, simple list of nodes that form the top view without climbing or guessing.

Before vs After

✗ Before

fn top_view_manual(tree: &TreeNode) {
    // Try to print nodes by guessing visibility
    // No clear way to track horizontal positions
    // May print duplicates or miss nodes
}

✓ After

use std::collections::{VecDeque, HashMap};

fn top_view(tree: &Option<Box<TreeNode>>) {
    // Use queue to do level order traversal
    // Track horizontal distance for each node
    // Print first node at each horizontal distance
}

What It Enables

This concept lets us quickly find which parts of a tree are visible from above, helping in visualization, map views, and solving complex tree problems easily.

Real Life Example

Think of a city skyline seen from a drone flying above. The top view of a binary tree is like identifying which buildings you see from above without any overlaps.

Key Takeaways

Manual checking of visible nodes is confusing and error-prone.

Top view uses horizontal distances and level order traversal to find visible nodes.

This method gives a clear, simple list of nodes visible from the top.

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