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

Time Complexity: Boundary Traversal of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time taken grows when we do a boundary traversal of a binary tree.

Specifically, how the number of nodes affects the work done to collect the boundary nodes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void printBoundary(Node* root) {
    if (!root) return;
    cout << root->data << " ";
    printLeftBoundary(root->left);
    printLeaves(root->left);
    printLeaves(root->right);
    printRightBoundary(root->right);
}

// Helper functions traverse parts of the tree
void printLeftBoundary(Node* node) {
    while (node) {
        if (!isLeaf(node)) cout << node->data << " ";
        if (node->left) node = node->left;
        else node = node->right;
    }
}

This code prints the boundary nodes of a binary tree in anti-clockwise order: root, left boundary, leaves, and right boundary.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing nodes in the tree to print left boundary, leaves, and right boundary.
• How many times: Each node is visited at most once during these traversals.

How Execution Grows With Input

As the number of nodes increases, the traversal visits each node once to check if it belongs to the boundary.

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

Pattern observation: The work grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The boundary traversal only visits boundary nodes, so it must be faster than visiting all nodes."

[OK] Correct: To find boundary nodes, the algorithm must check many nodes, including leaves and internal nodes, so it ends up visiting all nodes once.

Interview Connect

Understanding this traversal helps you reason about tree traversals and how to efficiently visit nodes in a specific order, a useful skill in many coding problems.

Self-Check

"What if we changed the tree to be skewed (all nodes only have one child)? How would the time complexity change?"

Practice

1. What does the boundary traversal of a binary tree include?

easy

A. Left boundary, leaves, and right boundary nodes in order

B. Only leaf nodes from left to right

C. All nodes in level order

D. Only root and leaf nodes

Solution

1. Step 1: Understand boundary traversal components

   Boundary traversal includes the left boundary (excluding leaves), all leaf nodes, and the right boundary (excluding leaves) in reverse order.

2. Step 2: Combine all parts in order

   Traverse left boundary top-down, then leaves left to right, then right boundary bottom-up to get the full boundary traversal.

3. Final Answer:

   Left boundary, leaves, and right boundary nodes in order -> Option A

4. Quick Check:

   Boundary traversal = Left boundary + Leaves + Right boundary [OK]

Hint: Remember: left boundary + leaves + right boundary [OK]

Common Mistakes:

• Including leaf nodes twice in left or right boundary
• Ignoring right boundary order (should be bottom-up)
• Confusing boundary traversal with level order

2. Which of the following C++ function signatures is correct for printing the left boundary of a binary tree?

easy

A. void printLeftBoundary(Node* root);

B. int printLeftBoundary(Node root);

C. void printLeftBoundary(Node root);

D. int printLeftBoundary(Node* root);

Solution

1. Step 1: Identify correct parameter type for tree node

   Binary tree nodes are passed as pointers (Node*), not by value (Node), to avoid copying the entire subtree.

2. Step 2: Check return type for printing function

   Printing functions usually return void as they output directly and do not return a value.

3. Final Answer:

   void printLeftBoundary(Node* root); -> Option A

4. Quick Check:

   Pointer parameter + void return = correct signature [OK]

Hint: Use pointer parameter and void return for print functions [OK]

Common Mistakes:

• Passing node by value causing unnecessary copies
• Using int return type for printing functions
• Missing pointer symbol (*) in parameter

3. Given the binary tree below, what is the boundary traversal output?

        1
       / \
      2   3
     / \   \
    4   5   6
       / \  /
      7  8 9

medium

A. [1, 2, 4, 7, 8, 9, 6]

B. [1, 2, 4, 7, 8, 9, 6, 3]

C. [1, 2, 4, 5, 7, 8, 9, 6, 3]

D. [1, 2, 4, 7, 8, 9, 3, 6]

Solution

1. Step 1: Identify left boundary nodes excluding leaves

   Left boundary: 1, 2 (4 is leaf, so excluded here)

2. Step 2: Collect all leaf nodes left to right

   Leaves: 4, 7, 8, 9

3. Step 3: Identify right boundary nodes excluding leaves in bottom-up order

   Right boundary: 6, 3

4. Step 4: Combine all parts

   Boundary traversal = [1, 2] + [4, 7, 8, 9] + [6, 3] = [1, 2, 4, 7, 8, 9, 6, 3]

5. Final Answer:

   [1, 2, 4, 7, 8, 9, 6, 3] -> Option B

6. Quick Check:

   Left boundary + leaves + right boundary = [1, 2, 4, 7, 8, 9, 6, 3] [OK]

Hint: Left boundary + leaves + right boundary in order [OK]

Common Mistakes:

• Including leaf nodes twice in boundary
• Wrong order of right boundary nodes
• Missing some leaf nodes

4. The following code snippet is intended to print the right boundary of a binary tree. What is the error?

void printRightBoundary(Node* root) {
    if (!root) return;
    if (root->right) {
        printRightBoundary(root->right);
    } else if (root->left) {
        printRightBoundary(root->left);
    }
    std::cout << root->data << " ";
}

medium

A. It misses printing the root node

B. It does not print nodes in bottom-up order

C. It prints leaf nodes which should be excluded

D. It causes infinite recursion

Solution

1. Step 1: Analyze printing condition in recursion

   The function prints every node after recursive calls, including leaf nodes which should be excluded from right boundary printing.

2. Step 2: Identify missing leaf node check

   There is no check to skip printing leaf nodes, so leaf nodes get printed incorrectly as part of right boundary.

3. Final Answer:

   It prints leaf nodes which should be excluded -> Option C

4. Quick Check:

   Leaf nodes must be excluded from right boundary [OK]

Hint: Skip printing leaf nodes in boundary functions [OK]

Common Mistakes:

• Printing leaf nodes in boundary traversal
• Not checking for null nodes before recursion
• Confusing order of printing nodes

5. You want to perform boundary traversal on a binary tree that may have only one node or be skewed (all left or all right). Which approach correctly handles these edge cases?

hard

A. Print only leaf nodes as boundary in skewed trees

B. Print all nodes in level order to cover all cases

C. Print left boundary including leaves, then right boundary including leaves, ignoring root

D. Print root, then left boundary excluding leaves, then leaves, then right boundary excluding leaves, skipping duplicates

Solution

1. Step 1: Understand edge cases in boundary traversal

   Single node tree or skewed trees may cause duplicates if leaves are included in boundaries without checks.

2. Step 2: Use approach that prints root, left boundary excluding leaves, leaves, right boundary excluding leaves, and skip duplicates

   This ensures no node is printed twice and all edge cases are handled correctly.

3. Final Answer:

   Print root, then left boundary excluding leaves, then leaves, then right boundary excluding leaves, skipping duplicates -> Option D

4. Quick Check:

   Exclude leaves in boundaries + skip duplicates [OK]

Hint: Exclude leaves in boundaries and skip duplicates [OK]

Common Mistakes:

• Printing leaves twice in skewed trees
• Ignoring root node in traversal
• Using level order which is not boundary traversal