Top View of Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a struct for a binary tree node.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* [1];
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Drag options to blanks, or click blank then click option'

Achild

Bparent

Cnext

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' instead of 'right' for the right child pointer.

Using 'next' which is common in linked lists but not binary trees.

The right child pointer in a binary tree node is commonly named 'right'.

2fill in blank

medium

Complete the code to insert the root node into the queue for level order traversal.

DSA C++

std::queue<std::pair<Node*, int>> q;
q.push([1], 0});

Drag options to blanks, or click blank then click option'

Aroot

Broot->data

Cnullptr

Dnew Node(0)

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing root->data instead of root pointer.

Pushing nullptr which causes runtime errors.

We push the root node pointer itself, not its data, into the queue for traversal.

3fill in blank

hard

Fix the error in the condition to check if a horizontal distance is not yet in the map.

DSA C++

if (topViewMap.find([1]) == topViewMap.end()) {
    topViewMap[hd] = node->data;
}

Drag options to blanks, or click blank then click option'

Ahd

Bnode->data

Cnode

DtopViewMap

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for node->data instead of hd in the map.

Using the map itself as a key.

We check if the horizontal distance 'hd' is in the map, not the node data.

4fill in blank

hard

Fill both blanks to correctly update the queue with left and right children and their horizontal distances.

DSA C++

if (node->left) q.push({node->left, hd [1] 1});
if (node->right) q.push({node->right, hd [2] 1});

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1 to left child hd or subtracting 1 from right child hd.

Using multiplication or division instead of addition/subtraction.

Left child horizontal distance is hd - 1; right child is hd + 1.

5fill in blank

hard

Fill all three blanks to create a map from horizontal distance to node data, and print the top view in order.

DSA C++

std::map<int, int> topViewMap;
// After filling topViewMap
for (auto [1] : topViewMap) {
    std::cout << [2] << ' ';
}
std::cout << std::endl;

// Function signature
std::vector<int> topView(Node* [3]) { /* ... */ }

Drag options to blanks, or click blank then click option'

Apair

Bsecond

Croot

Dfirst

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the key (first) instead of the value (second).

Using wrong variable names in the loop.

Wrong function parameter name.

We iterate over map pairs, print the second (node data), and the function takes root node pointer.

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