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

Time Complexity: Bottom View of Binary Tree

O(n log n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void bottomView(Node* root) {
    if (!root) return;
    map hdNodeMap; // horizontal distance to node data
    queue> q;
    q.push({root, 0});
    while (!q.empty()) {
        auto nodePair = q.front(); q.pop();
        Node* node = nodePair.first;
        int hd = nodePair.second;
        hdNodeMap[hd] = node->data; // overwrite for bottom view
        if (node->left) q.push({node->left, hd - 1});
        if (node->right) q.push({node->right, hd + 1});
    }
    for (auto& [hd, val] : hdNodeMap) {
        cout << val << " ";
    }
}
    

This code finds the bottom view of a binary tree by level order traversal and tracking horizontal distances.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node in the tree (n times).
• Inside the loop, updating the map is done for each node.

How Execution Grows With Input

As the number of nodes (n) increases, the number of times we visit nodes and update the map grows linearly.

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

Pattern observation: The work grows directly with the number of nodes, doubling nodes roughly doubles work.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit more than linearly because each node insertion into the map takes logarithmic time.

Common Mistake

[X] Wrong: "The time complexity is O(n) because we visit each node once."

[OK] Correct: Although each node is visited once, updating the map takes O(log n) time, so total time is O(n log n).

Interview Connect

Understanding how tree traversal combined with data structure operations affects time helps you explain your approach clearly and confidently.

Self-Check

"What if we used an unordered_map instead of map? How would the time complexity change?"

Practice

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

easy

A. Nodes visible when the tree is viewed from the bottom

B. Nodes visible when the tree is viewed from the top

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

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

Solution

1. Step 1: Understand the bottom view concept

   The bottom view shows nodes visible when looking from the bottom of the tree, meaning the lowest nodes at each horizontal distance.

2. Step 2: Compare with other views

   Top view shows nodes visible from above, left and right views show nodes from sides, so bottom view is unique for bottom visibility.

3. Final Answer:

   Nodes visible when the tree is viewed from the bottom -> Option A

4. Quick Check:

   Bottom view = nodes visible from bottom [OK]

Hint: Bottom view shows lowest nodes at each horizontal distance [OK]

Common Mistakes:

• Confusing bottom view with top view
• Thinking bottom view shows all leaf nodes
• Mixing bottom view with side views

2. Which of the following C++ STL containers is most suitable to store nodes by their horizontal distance for bottom view calculation?

easy

A. std::vector<int>

B. std::map

C. std::stack<int>

D. std::queue<int>

Solution

1. Step 1: Identify data structure needs

   We need to map horizontal distances (int) to node values (int) and keep them sorted by horizontal distance.

2. Step 2: Choose container

   std::map keeps keys sorted and allows key-value pairs, perfect for horizontal distance mapping.

3. Final Answer:

   std::map -> Option B

4. Quick Check:

   Use map for sorted key-value pairs [OK]

Hint: Use map to keep horizontal distances sorted [OK]

Common Mistakes:

• Using vector which is index-based, not key-based
• Using stack or queue which don't map keys to values
• Not sorting horizontal distances

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

        20
       /  \
     8     22
    / \      \
   5   3      25
      /  \    
     10   14

medium

A. [5, 8, 10, 14, 25]

B. [5, 10, 14, 22, 25]

C. [5, 10, 3, 14, 25]

D. [8, 10, 3, 14, 22]

Solution

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

   Root 20 at HD=0; 8 at HD=-1; 22 at HD=1; 5 at HD=-2; 3 at HD=0; 25 at HD=2; 10 at HD=-1; 14 at HD=1.

2. Step 2: Determine bottom view nodes by HD

   At HD=-2: 5; HD=-1: 10 (lower than 8); HD=0: 3 (lower than 20); HD=1: 14 (lower than 22); HD=2: 25.

3. Final Answer:

   [5, 10, 3, 14, 25] -> Option C

4. Quick Check:

   Bottom view nodes by HD = [5,10,3,14,25] [OK]

Hint: Track horizontal distances and pick lowest nodes [OK]

Common Mistakes:

• Choosing top nodes instead of bottom nodes
• Mixing horizontal distances incorrectly
• Ignoring nodes at same horizontal distance but lower level

4. Consider this code snippet for bottom view calculation. What is the error?

void bottomView(Node* root) {
  if (!root) return;
  std::map hdNodeMap;
  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 (hdNodeMap.find(hd) == hdNodeMap.end()) {
      hdNodeMap[hd] = node->data;
    }
    if (node->left) q.push({node->left, hd - 1});
    if (node->right) q.push({node->right, hd + 1});
  }
  for (auto it : hdNodeMap) {
    std::cout << it.second << " ";
  }
}

medium

A. It stores only the first node at each horizontal distance, not the bottom-most

B. It uses queue incorrectly for BFS traversal

C. It does not check for null root

D. It prints nodes in reverse horizontal order

Solution

1. Step 1: Analyze node storage logic

   The code stores node data only if horizontal distance is not already in map, so it keeps first node encountered at that HD.

2. Step 2: Understand bottom view requirement

   Bottom view requires the lowest node at each HD, so we must update map with nodes at lower levels, not just first.

3. Final Answer:

   It stores only the first node at each horizontal distance, not the bottom-most -> Option A

4. Quick Check:

   Bottom view needs last node at HD, not first [OK]

Hint: Update map with latest node at each horizontal distance [OK]

Common Mistakes:

• Not updating map for lower level nodes
• Assuming BFS order guarantees bottom view
• Ignoring horizontal distance updates

5. Given a binary tree, how would you modify the bottom view algorithm to handle nodes with the same horizontal distance and depth, ensuring the rightmost node is chosen in the bottom view?

hard

A. Use a queue for BFS and update map only if node depth is greater, replacing existing node

B. Use DFS and store only the first node encountered at each horizontal distance

C. Use a stack and update map only if node depth is less than current stored depth

D. Use BFS and update map only if node depth is greater or equal, replacing existing node

Solution

1. Step 1: Understand the problem with same HD and depth

   When nodes share horizontal distance and depth, bottom view should show the rightmost node.

2. Step 2: Modify BFS approach

   Use BFS to traverse level by level, and update map if current node's depth is greater or equal to stored node's depth, ensuring rightmost node replaces leftmost at same depth.

3. Step 3: Implement depth tracking

   Track depth along with horizontal distance, update map accordingly during BFS.

4. Final Answer:

   Use BFS and update map only if node depth is greater or equal, replacing existing node -> Option D

5. Quick Check:

   Update map on equal or greater depth for rightmost node [OK]

Hint: Update map if depth >= current to keep rightmost node [OK]

Common Mistakes:

• Ignoring depth when updating map
• Using DFS which doesn't guarantee level order
• Not handling equal depth nodes properly