Vertical Order Traversal of Binary Tree in DSA C++ - Time & Space Complexity

Time Complexity: Vertical Order Traversal of Binary Tree

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to do vertical order traversal changes as the tree grows.

How does the number of nodes affect the work done to group nodes by their vertical columns?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
  int data;
  Node* left;
  Node* right;
};

void verticalOrder(Node* root) {
  map<int, vector<int>> nodes;
  queue<pair<Node*, int>> q;
  q.push({root, 0});
  while (!q.empty()) {
    auto [node, hd] = q.front(); q.pop();
    nodes[hd].push_back(node->data);
    if (node->left) q.push({node->left, hd - 1});
    if (node->right) q.push({node->right, hd + 1});
  }
}
    

This code visits each node once, grouping nodes by their horizontal distance (hd) from the root.

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.

How Execution Grows With Input

As the number of nodes grows, the work grows roughly the same amount.

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

Pattern observation: The operations grow 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 of sorting keys in the map, but mostly it depends on the number of nodes.

Common Mistake

[X] Wrong: "The traversal is just O(n) because we visit each node once, so no extra cost."

[OK] Correct: The map stores nodes by horizontal distance and keeps keys sorted, so inserting each node involves a log n cost, making total time O(n log n).

Interview Connect

Understanding this complexity helps you explain how tree traversals combined with data structures like maps affect performance, a common topic in interviews.

Self-Check

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

Practice

1. What does vertical order traversal of a binary tree mean?

easy

A. Nodes are visited level by level from top to bottom

B. Nodes are visited in inorder (left, root, right)

C. Nodes are visited in preorder (root, left, right)

D. Nodes are grouped by their vertical columns from left to right

Solution

1. Step 1: Understand vertical order traversal concept

   Vertical order traversal groups nodes by their vertical columns, where nodes in the same vertical line share the same horizontal distance from the root.

2. Step 2: Compare with other traversals

   Level order visits nodes by depth, preorder and inorder are depth-first traversals, not grouped by vertical columns.

3. Final Answer:

   Nodes are grouped by their vertical columns from left to right -> Option D

4. Quick Check:

   Vertical order = group by vertical columns [OK]

Hint: Vertical order groups nodes by vertical columns [OK]

Common Mistakes:

• Confusing vertical order with level order traversal
• Thinking preorder or inorder is vertical order
• Ignoring horizontal distance concept

2. Which C++ data structure is best suited to store nodes grouped by their vertical columns during vertical order traversal?

easy

A. std::vector<int>

B. std::queue<int>

C. std::map

D. std::stack<int>

Solution

1. Step 1: Identify need to group nodes by vertical column

   We need a structure that maps vertical column indices (integers) to lists of node values.

2. Step 2: Choose appropriate C++ container

   std::map maps column indices to vectors of node values, maintaining sorted order of columns automatically.

3. Final Answer:

   std::map -> Option C

4. Quick Check:

   Map column index to node list = std::map [OK]

Hint: Use map to group nodes by column [OK]

Common Mistakes:

• Using vector alone loses column indexing
• Stack or queue do not map keys to values
• Using unordered_map loses column order

3. Given the binary tree:

    1
   / \
  2   3
 /     \
4       5

What is the vertical order traversal output?

medium

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

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

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

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

Solution

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

   Root 1 has HD=0; left child 2 has HD=-1; right child 3 has HD=1; 4 is left of 2 so HD=-2; 5 is right of 3 so HD=2.

2. Step 2: Group nodes by HD and list in order

   HD=-2: [4], HD=-1: [2], HD=0: [1], HD=1: [3], HD=2: [5]. Combine left to right: [4, 2, 1, 3, 5].

3. Final Answer:

   [4, 2, 1, 3, 5] -> Option A

4. Quick Check:

   Vertical order = [4, 2, 1, 3, 5] [OK]

Hint: Assign horizontal distances, then list nodes left to right [OK]

Common Mistakes:

• Mixing order of nodes in same vertical column
• Ignoring horizontal distance for children
• Listing nodes by level order instead

4. Consider this code snippet for vertical order traversal:

void verticalOrder(Node* root) {
  map m;
  queue> q;
  q.push({root, 0});
  while (!q.empty()) {
    auto p = q.front(); q.pop();
    Node* node = p.first;
    int hd = p.second;
    m[hd].push_back(node->data);
    if (node->left) q.push({node->left, hd - 1});
    if (node->right) q.push({node->right, hd + 1});
  }
  for (auto &x : m) {
    for (int val : x.second) cout << val << " ";
  }
}

What is the bug in this code?

medium

A. It does not handle empty tree (root == nullptr)

B. It uses map instead of unordered_map causing wrong order

C. It pushes right child before left child in queue

D. It does not store horizontal distances correctly

Solution

1. Step 1: Check for null root before processing

   The code assumes root is not null but does not check. If root is nullptr, q.push({root, 0}) pushes null pointer causing runtime error.

2. Step 2: Verify other parts are correct

   Using map keeps columns sorted, order of pushing left then right is correct, horizontal distances are updated properly.

3. Final Answer:

   It does not handle empty tree (root == nullptr) -> Option A

4. Quick Check:

   Missing null root check causes error [OK]

Hint: Always check if root is nullptr before traversal [OK]

Common Mistakes:

• Ignoring null root causing crash
• Confusing map with unordered_map for order
• Swapping left and right child pushes

5. You want to modify vertical order traversal to print nodes sorted by their depth within each vertical column (top to bottom). Which data structure change helps achieve this?

hard

A. Use map and sort vectors after traversal

B. Use map> to store nodes by column and depth

C. Use unordered_map and sort keys later

D. Use queue> only without map

Solution

1. Step 1: Understand requirement to sort nodes by depth per column

   We need to group nodes by vertical column and also by their depth (level) to print top to bottom.

2. Step 2: Choose data structure to store nodes by column and depth

   Using map> allows mapping column -> depth -> list of nodes, preserving order in both dimensions.

3. Final Answer:

   Use map> to store nodes by column and depth -> Option B

4. Quick Check:

   Nested map for column and depth sorting [OK]

Hint: Use nested map to sort nodes by column and depth [OK]

Common Mistakes:

• Sorting vectors after traversal is inefficient
• Using unordered_map loses order
• Ignoring depth dimension in grouping