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 grows when we find the bottom view of a binary tree.

How does the number of steps change as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public IList<int> BottomView(TreeNode root) {
    var result = new List<int>();
    if (root == null) return result;

    var queue = new Queue<(TreeNode node, int hd)>();
    var map = new SortedDictionary<int, int>();
    queue.Enqueue((root, 0));

    while (queue.Count > 0) {
        var (node, hd) = queue.Dequeue();
        map[hd] = node.val; // overwrite to keep bottom-most
        if (node.left != null) queue.Enqueue((node.left, hd - 1));
        if (node.right != null) queue.Enqueue((node.right, hd + 1));
    }

    foreach (var val in map.Values) result.Add(val);
    return result;
}
    

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, so n times for n nodes.
• Additional operations: Inserting and updating entries in the map for each node.
• Final loop: Iterates over map entries, at most n entries.

How Execution Grows With Input

As the tree grows, the number of nodes increases, so the steps grow roughly the same.

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 linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit more than linearly because each map update takes extra time to keep keys sorted.

Common Mistake

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

[OK] Correct: Updating the sorted map for each node takes extra time, so the total is more than just visiting nodes.

Interview Connect

Understanding how traversal and data structure choices affect time helps you explain your approach clearly and confidently.

Self-Check

"What if we used a hash map instead of a sorted map? How would the time complexity change?"

Practice

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

easy

A. The nodes visible when the tree is viewed from the right side

B. The nodes visible when the tree is viewed from the top

C. The nodes visible when the tree is viewed from the left side

D. The nodes visible when the tree is viewed from the bottom

Solution

1. Step 1: Understand the bottom view concept

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

2. Step 2: Compare with other views

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

3. Final Answer:

   The nodes visible when the tree is viewed from the bottom -> Option D

4. Quick Check:

   Bottom view = nodes visible from below [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 C# data structure is best suited to store nodes by their horizontal distance when computing the bottom view of a binary tree?

easy

A. Dictionary

B. List<int>

C. Stack<int>

D. Queue<int>

Solution

1. Step 1: Identify the need to map horizontal distance to node value

   We need a structure to store nodes keyed by their horizontal distance for quick updates and lookups.

2. Step 2: Choose the appropriate data structure

   A Dictionary allows mapping horizontal distance (int) to node value (int) efficiently.

3. Final Answer:

   Dictionary<int, int> -> Option A

4. Quick Check:

   Use Dictionary for horizontal distance mapping [OK]

Hint: Use Dictionary to map horizontal distance to node value [OK]

Common Mistakes:

• Using List which doesn't map keys to values
• Using Stack or Queue which are not key-value stores
• Confusing data structure roles

3. Consider the following binary tree:

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

What is the bottom view of this tree?

medium

A. [5, 10, 3, 22, 25]

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

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

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

Solution

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

   Assign horizontal distances (HD) starting from root (20) at HD=0: - 20(0), 8(-1), 22(1), 5(-2), 3(0), 25(2), 10(-1), 14(1) For each HD, the bottom-most node is: HD -2: 5 HD -1: 10 (overwrites 8) HD 0: 3 (overwrites 20) HD 1: 14 (overwrites 22) HD 2: 25

2. Step 2: Collect bottom view nodes in order of HD

   Order HD from left to right: -2, -1, 0, 1, 2 Bottom view nodes: [5, 10, 3, 14, 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 distance and overwrite with lower nodes [OK]

Common Mistakes:

• Not updating nodes at same horizontal distance
• Confusing top view with bottom view
• Ignoring horizontal distance ordering

4. The following C# code attempts to compute the bottom view of a binary tree but has a bug. What is the error?

void BottomView(Node root) {
    if (root == null) return;
    var map = new Dictionary();
    var queue = new Queue<(Node node, int hd)>();
    queue.Enqueue((root, 0));
    while (queue.Count > 0) {
        var (node, hd) = queue.Dequeue();
        if (!map.ContainsKey(hd)) {
            map[hd] = node.data;
        }
        if (node.left != null) queue.Enqueue((node.left, hd - 1));
        if (node.right != null) queue.Enqueue((node.right, hd + 1));
    }
    foreach (var key in map.Keys.OrderBy(k => k)) {
        Console.Write(map[key] + " ");
    }
}

medium

A. It does not enqueue the root node

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

C. It uses a queue instead of a stack

D. It does not order the keys before printing

Solution

1. Step 1: Analyze how nodes are stored in the map

   The code adds node data to map only if the horizontal distance key is not present, so it keeps the first node encountered at that HD.

2. Step 2: Understand bottom view requires last node at each HD

   Bottom view needs the lowest node at each HD, so the map should update the value every time a node at the same HD is found, not only the first.

3. Final Answer:

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

4. Quick Check:

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

Hint: Update map value for each node at HD to get bottom view [OK]

Common Mistakes:

• Using if (!map.ContainsKey) instead of always updating
• Ignoring horizontal distance updates
• Not ordering keys before printing (though code does)

5. You are given a binary tree where multiple nodes share the same horizontal distance and depth. How would you modify the bottom view algorithm to always show the node with the greatest depth at each horizontal distance?

hard

A. Store pairs of (node value, depth) in the map and update only if new node's depth is greater

B. Store only node values and update map only if node value is greater

C. Use a stack instead of a queue to traverse nodes

D. Ignore depth and just print nodes in level order

Solution

1. Step 1: Recognize the need to track depth with node values

   To ensure the bottom view shows the node with greatest depth at each horizontal distance, we must store depth along with node value.

2. Step 2: Update map only if new node's depth is greater

   When visiting a node, compare its depth with stored depth for that horizontal distance; update only if current node is deeper.

3. Final Answer:

   Store pairs of (node value, depth) in the map and update only if new node's depth is greater -> Option A

4. Quick Check:

   Track depth and update map for deeper nodes [OK]

Hint: Keep depth info and update map only if deeper node found [OK]

Common Mistakes:

• Ignoring depth and overwriting incorrectly
• Comparing node values instead of depths
• Changing traversal order without depth tracking