Why Top View of Binary Tree in DSA C++?

The Big Idea

What if you could instantly see the outline of a complex tree without checking every node?

The Scenario

Imagine you have a tall building with many floors and rooms. You want to see which rooms are visible if you look at the building from the front, ignoring the rooms behind others. Doing this by checking each room one by one from the front is very hard.

The Problem

Manually checking each room from the front means you might miss some rooms hidden behind others or spend too much time looking at every floor and room. It's easy to get confused and make mistakes because you have to remember which rooms are in front and which are behind.

The Solution

The top view of a binary tree shows exactly which nodes are visible when you look from above. Using a smart method, we can find these nodes quickly by tracking their horizontal positions and picking the first node seen at each position. This saves time and avoids confusion.

Before vs After

✗ Before

void printTopView(Node* root) {
  // Manually check each node and compare positions
  // Very complex and slow
}

✓ After

void printTopView(Node* root) {
  map<int, int> topNodes;
  queue<pair<Node*, int>> nodesQueue;
  // Efficiently track and print top view
}

What It Enables

This lets you quickly see the outline of a tree from above, which helps in understanding its shape and structure easily.

Real Life Example

In city planning, seeing the skyline of buildings from above helps decide where to build new roads or parks. Similarly, the top view of a tree helps in visualizing data structures in computer science.

Key Takeaways

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

Top view uses horizontal positions to find visible nodes efficiently.

This method helps visualize the tree's shape from above clearly.

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