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

Time Complexity: Top View of Binary Tree

O(n log 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.

// Function to print top view of 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 << " ";
}
    

This code uses a breadth-first search (BFS) to visit nodes level by level and records the first node encountered at each horizontal distance (hd) using a map.

Identify Repeating Operations

• Primary loop: BFS while loop visits each of the n nodes exactly once.
• Per-node operations: Queue pop (O(1)), map find (O(log w), where w is current map size ≤ n), conditional insert (O(log n) at most once per unique hd).
• Key insight: Map operations (find/insert) occur for every node, not just unique hd.

How Execution Grows With Input

The BFS visits n nodes, but each visit includes O(log n) map work. Total work: O(n log n). Printing traverses the map once: O(w) where w ≤ n (negligible).

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 30 map ops + queue work
100	About 100 * log(100) ≈ 700 map ops
1000	About 1000 * log(1000) ≈ 10,000 map ops

Pattern observation: Grows as n log n, faster than linear due to map lookups per node.

Final Time Complexity

Time Complexity: O(n log n)

Dominated by n map operations, each O(log n). Space: O(w) for map + O(w) queue, worst O(n).

Common Mistake

[X] Wrong: "Visits each node once, map only inserts unique hd (≤ n), so O(n)."

[OK] Correct: Map find runs for every node (n times), each O(log n), even if insert skipped. Total: O(n log n).

Interview Connect

Spotting the hidden log n from map usage per node distinguishes solid analysis. Optimize with unordered_map (avg O(n)) or vertical order traversal tweaks.

Self-Check

"What if we used unordered_map instead of map? How would the time complexity change in average/worst case?"

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