Top View of Binary Tree in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Top View for a Simple Binary Tree

What is the output of the top view traversal for the given binary tree?

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

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

// Top view expected: 2 -> 1 -> 3 -> 5

#include <iostream>
#include <map>
#include <queue>

void topView(Node* root) {
    if (!root) return;
    std::map<int, int> topNode;
    std::queue<std::pair<Node*, int>> q;
    q.push({root, 0});
    while (!q.empty()) {
        auto p = q.front(); q.pop();
        Node* node = p.first;
        int hd = p.second;
        if (topNode.find(hd) == topNode.end()) {
            topNode[hd] = node->data;
        }
        if (node->left) q.push({node->left, hd - 1});
        if (node->right) q.push({node->right, hd + 1});
    }
    for (auto it : topNode) {
        std::cout << it.second << " -> ";
    }
    std::cout << "null" << std::endl;
}

int main() {
    Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->right = new Node(4);
    root->right->right = new Node(5);
    topView(root);
    return 0;
}

A 4 -> 2 -> 1 -> 3 -> null

B 2 -> 1 -> 3 -> 5 -> null

C 2 -> 4 -> 1 -> 3 -> 5 -> null

D 1 -> 2 -> 3 -> 4 -> 5 -> null

💡 Hint

Remember top view shows nodes visible from above, one per horizontal distance.

The top view prints the first node encountered at each horizontal distance from the root.

Here, nodes 2, 1, 3, and 5 are visible from the top at horizontal distances -1, 0, 1, and 2 respectively.

🧠 Conceptual

intermediate

Horizontal Distance in Top View

In the top view of a binary tree, what does the horizontal distance (hd) represent?

A The distance of a node from the root measured horizontally, left is negative, right is positive

B The vertical level of a node from the root

C The depth of the node in the tree

D The number of nodes in the subtree rooted at that node

💡 Hint

Think about how nodes are positioned left or right relative to the root.

Horizontal distance is used to track how far left or right a node is from the root.

Left child decreases hd by 1, right child increases hd by 1.

🔧 Debug

advanced

Identify the Bug in Top View Implementation

What error will occur when running this top view code snippet?

DSA C++

void topView(Node* root) {
    if (!root) return;
    std::map<int, int> topNode;
    std::queue<std::pair<Node*, int>> q;
    q.push({root, 0});
    while (!q.empty()) {
        auto p = q.front(); q.pop();
        Node* node = p.first;
        int hd = p.second;
        topNode[hd] = node->data; // Bug here: overwrites existing
        if (node->left) q.push({node->left, hd - 1});
        if (node->right) q.push({node->right, hd + 1});
    }
    for (auto it : topNode) {
        std::cout << it.second << " -> ";
    }
    std::cout << "null" << std::endl;
}

A Runtime error due to null pointer dereference

B Compilation error due to incorrect queue usage

C The output will show only the last node at each horizontal distance, not the top view

D No error, output is correct top view

💡 Hint

Check how the map is updated inside the loop.

The code overwrites the value for each horizontal distance every time a node is visited.

This causes the bottom-most node at that horizontal distance to be stored, not the top-most.

🚀 Application

advanced

Top View of a Complex Binary Tree

Given the binary tree below, what is the top view output? // Tree structure: // 10 // / \ // 5 15 // / \ \ // 3 7 18 // \ / // 4 16 // Expected output format: node values separated by ' -> ' ending with 'null'

A 3 -> 5 -> 10 -> 15 -> 18 -> null

B 3 -> 4 -> 5 -> 10 -> 15 -> 16 -> 18 -> null

C 3 -> 5 -> 7 -> 10 -> 15 -> 18 -> null

D 4 -> 5 -> 10 -> 15 -> 18 -> null

💡 Hint

Remember top view shows the first node at each horizontal distance from left to right.

Horizontal distances and first nodes encountered are:

• hd -2: 3
• hd -1: 5
• hd 0: 10
• hd 1: 15
• hd 2: 18

Nodes 4, 7, and 16 are hidden behind others from top view.

🧠 Conceptual

expert

Why Use a Map in Top View Algorithm?

Why is a map (or dictionary) used to store nodes by horizontal distance in the top view algorithm?

A To store nodes by their depth level instead of horizontal distance

B To store all nodes at each horizontal distance for later processing

C To count the number of nodes at each horizontal distance

D To keep nodes sorted by their horizontal distance for ordered output

💡 Hint

Think about how the output order is determined.

The map keeps keys (horizontal distances) sorted automatically.

This allows printing the top view nodes from leftmost to rightmost in order.

Practice

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

easy

A. All leaf nodes of the tree

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

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

D. Nodes visible when the tree is viewed from above

Solution

1. Step 1: Understand the meaning of top view

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

2. Step 2: Compare with other views

   Left or right views show nodes visible from sides, and leaf nodes are only the bottom nodes, so they are different.

3. Final Answer:

   Nodes visible when the tree is viewed from above -> Option D

4. Quick Check:

   Top view = nodes seen from above [OK]

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

Common Mistakes:

• Confusing top view with left or right view
• Thinking top view shows all leaf nodes
• Assuming top view includes hidden nodes

2. Which C++ STL container is best suited to store nodes by their horizontal distance for top view calculation?

easy

A. std::map

B. std::stack

C. std::vector

D. std::queue

Solution

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

   Top view requires storing nodes by their horizontal distance from root, which can be negative or positive keys.

2. Step 2: Choose container supporting key-value pairs with sorted keys

   std::map stores keys in sorted order and allows negative keys, unlike vector which uses indices starting at 0.

3. Final Answer:

   std::map -> Option A

4. Quick Check:

   Use map for horizontal distance keys [OK]

Hint: Use map to store nodes by horizontal distance [OK]

Common Mistakes:

• Using vector which can't handle negative keys
• Using stack or queue which don't map keys
• Not sorting nodes by horizontal distance

3. What is the output of the top view for the following binary tree?

        1
       / \
      2   3
       \   \
        4   5
         \   \
          6   7

medium

A. [2, 1, 3, 5, 7]

B. [2, 1, 4, 6, 7]

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

D. [1, 2, 3, 4, 5, 6, 7]

Solution

1. Step 1: Assign horizontal distances (HD) to nodes

   Root 1 has HD=0; left child 2 is HD=-1; right child 3 is HD=1; 4 is right of 2 so HD=0; 5 is right of 3 so HD=2; 6 is right of 4 so HD=1; 7 is right of 5 so HD=3.

2. Step 2: Select first node at each HD from top to bottom

   HD=-1: 2; HD=0: 1; HD=1: 3; HD=2: 5; HD=3: 7. Nodes 4 and 6 are hidden behind 1 and 3 respectively.

3. Final Answer:

   [2, 1, 3, 5, 7] -> Option A

4. Quick Check:

   Top view nodes by HD = [2,1,3,5,7] [OK]

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

Common Mistakes:

• Including nodes hidden behind others at same HD
• Mixing order of nodes in output
• Ignoring horizontal distance calculation

4. Identify the error in this C++ snippet for top view calculation:

std::map topViewMap;
std::queue> q;
q.push({root, 0});
while(!q.empty()) {
  auto p = q.front(); q.pop();
  Node* node = p.first;
  int hd = p.second;
  if(topViewMap.find(hd) == topViewMap.end())
    topViewMap[hd] = node->data;
  if(node->left) q.push({node->left, hd - 1});
  if(node->right) q.push({node->right, hd + 1});
}

medium

A. Using std::map instead of unordered_map causes slower lookup

B. Incorrect horizontal distance update for left and right children

C. Not checking if node is nullptr before accessing node->left and node->right

D. Not clearing the queue before starting the loop

Solution

1. Step 1: Check node pointer usage inside loop

   The code accesses node->left and node->right without verifying if node is nullptr, which can cause runtime errors if node is null.

2. Step 2: Verify other parts of code

   Horizontal distances are updated correctly; queue is initialized properly; map choice is valid though unordered_map could be faster but not an error.

3. Final Answer:

   Not checking if node is nullptr before accessing node->left and node->right -> Option C

4. Quick Check:

   Always check node != nullptr before child access [OK]

Hint: Check node pointer before accessing children [OK]

Common Mistakes:

• Ignoring nullptr checks causing crashes
• Confusing horizontal distance updates
• Assuming map choice causes errors

5. Given a binary tree where multiple nodes share the same horizontal distance and level, which node should appear in the top view?

hard

A. The node with the maximum value at that horizontal distance

B. The node that appears first in level order traversal at that horizontal distance

C. The node that appears last in level order traversal at that horizontal distance

D. The leftmost node at that horizontal distance regardless of level

Solution

1. Step 1: Understand top view node selection criteria

   Top view shows the first node encountered at each horizontal distance when traversing level by level (top to bottom, left to right).

2. Step 2: Clarify why first node in level order is chosen

   Because level order visits nodes by increasing depth, the first node at a horizontal distance is the topmost visible node.

3. Final Answer:

   The node that appears first in level order traversal at that horizontal distance -> Option B

4. Quick Check:

   Top view = first node per horizontal distance in level order [OK]

Hint: Pick first node at horizontal distance in level order [OK]

Common Mistakes:

• Choosing node by max value instead of position
• Picking last node instead of first in level order
• Ignoring level order traversal importance