Top View of Binary Tree in DSA C - Time & Space Complexity

Time Complexity: Top View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the top view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done to get the top view?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Structure for tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};

// Function to print top view
void topView(struct Node* root) {
    if (!root) return;
    // Use a queue for level order traversal
    // Use a map to store first node at each horizontal distance
    // Traverse tree and print top view nodes
}
    

This code performs a level order traversal to find the top view nodes of the binary tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Level order traversal visiting each node once.
• How many times: Each node is visited exactly one time.

How Execution Grows With Input

As the number of nodes increases, the traversal visits each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the top view grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The top view requires checking all paths multiple times, so it must be more than linear."

[OK] Correct: The level order traversal visits each node once, and the map stores only the first node at each horizontal distance, so no repeated work happens.

Interview Connect

Understanding this helps you explain how tree traversals work efficiently and how to use extra data structures to track needed information.

Self-Check

"What if we used a recursive approach instead of level order traversal? How would the time complexity change?"

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 right side

D. All leaf nodes of the tree

Solution

1. Step 1: Understand the meaning of top view

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

2. Step 2: Compare with other views

   Left or right views show nodes from side angles, and leaf nodes are only the bottom nodes, not the top view.

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 top view includes all leaf nodes
• Assuming top view shows all nodes

2. Which data structure is commonly used to find the top view of a binary tree in C?

easy

A. Stack

B. Queue

C. Hash Map

D. Linked List

Solution

1. Step 1: Identify traversal method for top view

   Top view is found using level order traversal, which uses a queue to process nodes level by level.

2. Step 2: Understand role of queue

   Queue helps visit nodes in order from top to bottom and left to right, needed for correct top view.

3. Final Answer:

   Queue -> Option B

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level order traversal in top view [OK]

Common Mistakes:

• Using stack which is for depth-first traversal
• Confusing hash map as traversal structure
• Using linked list instead of queue

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

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

medium

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

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

C. [2, 1, 3, 4, 7]

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

Solution

1. Step 1: Assign horizontal distances

   Assign horizontal distance (HD) from root: root(1) = 0, left child decreases HD by 1, right child increases HD by 1. So, 2 = -1, 3 = 1, 4 = 0, 5 = 2, 6 = 1, 7 = 3.

2. Step 2: Select topmost nodes at each HD

   At HD -1: 2; HD 0: 1 (4 is lower); HD 1: 3 (6 is lower); HD 2: 5; HD 3: 7.

3. Final Answer:

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

4. Quick Check:

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

Hint: Use horizontal distance and pick top nodes [OK]

Common Mistakes:

• Choosing lower nodes instead of top nodes
• Ignoring horizontal distance concept
• Mixing order of nodes in output

4. Consider this C code snippet to print top view of a binary tree:

void topView(Node* root) {
    if (!root) return;
    map topNode;
    queue> 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)
        cout << it.second << " ";
}

What is the error in this code?

medium

A. Incorrect horizontal distance calculation

B. Not checking if root is NULL

C. Using C++ STL map and queue in C code

D. Missing return statement in void function

Solution

1. Step 1: Identify language mismatch

   The code uses C++ STL containers like map and queue, but the function is declared in C style which does not support these STL features.

2. Step 2: Check other code parts

   Root NULL check is present, horizontal distance is correctly updated, and void functions do not require return statements.

3. Final Answer:

   Using C++ STL map and queue in C code -> Option C

4. Quick Check:

   C code cannot use C++ STL containers [OK]

Hint: C code can't use C++ STL containers like map or queue [OK]

Common Mistakes:

• Ignoring language differences between C and C++
• Thinking void functions need return
• Assuming horizontal distance logic is wrong

5. You have a binary tree where multiple nodes share the same horizontal distance and level. How should you modify the top view algorithm to always show the leftmost node at that horizontal distance?

hard

A. Store the first node encountered at each horizontal distance during level order traversal

B. Store the last node encountered at each horizontal distance during level order traversal

C. Store nodes with maximum depth at each horizontal distance

D. Store nodes with minimum depth at each horizontal distance but prefer right child first

Solution

1. Step 1: Understand node selection in top view

   Top view shows the first node encountered at each horizontal distance during level order traversal, which naturally picks the leftmost node at that distance.

2. Step 2: Avoid overwriting nodes

   By storing only the first node at each horizontal distance and ignoring later nodes, the leftmost node is preserved.

3. Final Answer:

   Store the first node encountered at each horizontal distance during level order traversal -> Option A

4. Quick Check:

   First node at HD = leftmost node [OK]

Hint: Keep first node per horizontal distance in level order [OK]

Common Mistakes:

• Overwriting with last node instead of first
• Choosing nodes by max depth instead of min
• Preferring right child first breaks leftmost rule