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

Time Complexity: Bottom View of Binary Tree

O(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.

Specifically, how does the work increase when the number of nodes increases?

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 bottom view
void bottomView(struct Node* root) {
    if (!root) return;
    // Use queue for level order traversal
    // Use map to store horizontal distance and node data
    // Traverse tree and update map
    // Print map values
}
    

This code performs a level order traversal of the tree, tracking horizontal distances to find the bottom view nodes.

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 processed 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 increase linearly as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the bottom view grows directly in proportion to the number of nodes in the tree.

Common Mistake

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

[OK] Correct: The level order traversal visits each node once, and updating the map is constant time per node, so total time stays linear.

Interview Connect

Understanding how tree traversals scale helps you explain your approach clearly and confidently in interviews.

Self-Check

"What if we used a recursive depth-first traversal instead of level order? 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 that are visible when looking at the tree from below, ignoring nodes hidden behind others.

2. Step 2: Compare with other views

   Top, left, and right views show nodes from different angles, but bottom view specifically shows the lowest nodes at each horizontal distance.

3. Final Answer:

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

4. Quick Check:

   Bottom view = nodes visible from below [OK]

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

Common Mistakes:

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

2. Which data structure is commonly used to track nodes and their horizontal distances for bottom view calculation?

easy

A. Stack

B. Hash Map (Dictionary)

C. Linked List

D. Queue

Solution

1. Step 1: Identify the need for horizontal distance mapping

   To find the bottom view, we map each node's horizontal distance to its value, updating with the lowest node at that distance.

2. Step 2: Choose the suitable data structure

   A hash map (dictionary) stores horizontal distance as key and node value as value, allowing quick updates and lookups.

3. Final Answer:

   Hash Map (Dictionary) -> Option B

4. Quick Check:

   Use hash map to map horizontal distance to node [OK]

Hint: Use hash map to store horizontal distance and node value [OK]

Common Mistakes:

• Using stack which is LIFO and not suitable here
• Using queue alone without mapping distances
• Using linked list which is inefficient for lookups

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

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

Options show node values from left to right in bottom view.

medium

A. [2, 4, 6, 7]

B. [2, 4, 6, 3, 7]

C. [2, 4, 6, 5, 7]

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

Solution

1. Step 1: Assign horizontal distances and track bottom nodes

   Assign horizontal distances: 1=0, 2=-1, 3=1, 4=0, 5=2, 6=1, 7=3. Bottom nodes at each distance are: -1:2, 0:4, 1:6, 2:5, 3:7.

2. Step 2: Collect bottom view nodes from left to right

   From leftmost (-1) to rightmost (3), nodes are 2, 4, 6, 5, 7.

3. Final Answer:

   [2, 4, 6, 5, 7] -> Option C

4. Quick Check:

   Bottom view nodes left to right = [2, 4, 6, 5, 7] [OK]

Hint: Track horizontal distances and pick last node at each [OK]

Common Mistakes:

• Including nodes not visible from bottom
• Confusing horizontal distances
• Listing nodes in wrong order

4. Identify the error in this C code snippet for bottom view calculation:

void bottomView(Node* root) {
    map hdNodeMap;
    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) {
        cout << it.second << " ";
    }
}

medium

A. Output loop should print keys instead of values

B. Queue is not initialized properly

C. Left and right child checks are reversed

D. Using map prevents updating bottom nodes at same horizontal distance

Solution

1. Step 1: Analyze map usage for horizontal distances

   The code inserts node data only if horizontal distance is not present, so it keeps the first node seen, which is top view behavior.

2. Step 2: Correct approach for bottom view

   For bottom view, update the map every time a node at the same horizontal distance is found, so the last node at that distance remains.

3. Final Answer:

   Using map prevents updating bottom nodes at same horizontal distance -> Option D

4. Quick Check:

   Bottom view needs updating map entries [OK]

Hint: Update map for every node at horizontal distance for bottom view [OK]

Common Mistakes:

• Using if condition to insert only once (top view logic)
• Not updating map with lower nodes
• Confusing top and bottom view logic

5. Given a binary tree where multiple nodes share the same horizontal distance and level, how should the bottom view algorithm decide which node to show?

hard

A. Show the node that appears last in level order traversal at that horizontal distance

B. Show the node that appears first in level order traversal at that horizontal distance

C. Show the node with the smallest value at that horizontal distance

D. Show the node with the largest value at that horizontal distance

Solution

1. Step 1: Understand node conflicts at same horizontal distance and level

   When nodes share horizontal distance and level, bottom view shows the one visible from bottom, which is the last node encountered in level order at that position.

2. Step 2: Use level order traversal order to decide

   Level order traversal visits nodes top to bottom, left to right. The last node at a horizontal distance overwrites previous ones in the map.

3. Final Answer:

   Show the node that appears last in level order traversal at that horizontal distance -> Option A

4. Quick Check:

   Bottom view picks last node at horizontal distance [OK]

Hint: Last node at horizontal distance in level order is bottom view node [OK]

Common Mistakes:

• Picking first node instead of last
• Choosing node by value instead of position
• Ignoring traversal order impact