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

Time Complexity: Boundary Traversal of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to do a boundary traversal of a binary tree changes as the tree grows.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function boundaryTraversal(root) {
  if (!root) return [];
  const result = [];
  function leftBoundary(node) {
    if (!node || (!node.left && !node.right)) return;
    result.push(node.val);
    if (node.left) leftBoundary(node.left);
    else leftBoundary(node.right);
  }
  function leaves(node) {
    if (!node) return;
    leaves(node.left);
    if (!node.left && !node.right) result.push(node.val);
    leaves(node.right);
  }
  function rightBoundary(node) {
    if (!node || (!node.left && !node.right)) return;
    if (node.right) rightBoundary(node.right);
    else rightBoundary(node.left);
    result.push(node.val);
  }
  result.push(root.val);
  leftBoundary(root.left);
  leaves(root.left);
  leaves(root.right);
  rightBoundary(root.right);
  return result;
}
    

This code collects the boundary nodes of a binary tree in order: left edge, leaves, then right edge.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive traversal of nodes in the tree.
• How many times: The tree nodes are visited a constant number of times each (at most twice).

How Execution Grows With Input

As the number of nodes (n) increases, the total number of node visits is proportional to n, so the work grows linearly with n.

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

Pattern observation: The number of operations grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to do the boundary traversal grows in direct proportion to 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: The code still visits many nodes to find leaves and edges, so it ends up visiting every node in the tree.

Interview Connect

Understanding how tree traversals scale helps you explain your approach clearly and shows you know how to handle data structures efficiently.

Self-Check

"What if we changed the code to use iterative traversal with stacks instead of recursion? How would the time complexity change?"

Practice

1. What is the main purpose of Boundary Traversal in a binary tree?

easy

A. To print all leaf nodes only

B. To print all nodes on the boundary in an anti-clockwise direction

C. To print nodes level by level

D. To print nodes in in-order traversal

Solution

1. Step 1: Understand boundary traversal definition

   Boundary traversal prints nodes on the edges of the tree, including left boundary, leaves, and right boundary.

2. Step 2: Compare options with definition

   Only To print all nodes on the boundary in an anti-clockwise direction describes printing all boundary nodes in anti-clockwise order.

3. Final Answer:

   To print all nodes on the boundary in an anti-clockwise direction -> Option B

4. Quick Check:

   Boundary traversal = print boundary nodes anti-clockwise [OK]

Hint: Boundary traversal prints edges and leaves anti-clockwise [OK]

Common Mistakes:

• Confusing boundary traversal with leaf node traversal
• Thinking it prints nodes level-wise
• Mixing it with in-order traversal

2. Which of the following is the correct way to start the boundary traversal function in JavaScript?

easy

A. function boundaryTraversal(root) { console.log(root.right); }

B. function boundaryTraversal() { return root; }

C. function boundaryTraversal(root) { if (!root) return []; }

D. function boundaryTraversal(root) { return root.left; }

Solution

1. Step 1: Check function signature and base case

   The function should accept root and handle empty tree by returning empty array.

2. Step 2: Validate options

   function boundaryTraversal(root) { if (!root) return []; } correctly defines function with root parameter and returns empty array if root is null.

3. Final Answer:

   function boundaryTraversal(root) { if (!root) return []; } -> Option C

4. Quick Check:

   Function starts with root check and empty return [OK]

Hint: Always check for empty tree at start of traversal function [OK]

Common Mistakes:

• Omitting root parameter
• Not handling empty tree
• Returning wrong values at start

3. Given the binary tree:

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

What is the output of boundary traversal (anti-clockwise)?

medium

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

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

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

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

Solution

1. Step 1: Identify left boundary nodes

   Left boundary (excluding leaves): 1, 2

2. Step 2: Identify leaf nodes left to right

   Leaves: 4, 7, 8, 9, 6

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

   Right boundary: 3

4. Step 4: Combine in anti-clockwise order

   Boundary traversal: 1 (root), 2 (left boundary), 4 (leaf), 7, 8, 9, 6 (leaf), 3 (right boundary)

5. Final Answer:

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

6. Quick Check:

   Boundary nodes combined anti-clockwise = [1, 2, 4, 7, 8, 9, 6, 3] [OK]

Hint: List left boundary, leaves, then right boundary reversed [OK]

Common Mistakes:

• Including internal nodes as leaves
• Adding right boundary leaves twice
• Wrong order of nodes

4. Consider this code snippet for boundary traversal in JavaScript:

function printLeaves(node, result) {
  if (!node) return;
  if (!node.left && !node.right) {
    result.push(node.data);
  }
  printLeaves(node.left, result);
  printLeaves(node.right, result);
}

What is the bug in this code?

medium

A. It visits leaves in left to right order correctly

B. It pushes non-leaf nodes into result

C. It visits leaves in wrong order (right before left)

D. It does not check for null node before recursion

Solution

1. Step 1: Analyze leaf detection and recursion order

   Code checks if node is leaf and pushes data, then recursively visits left then right child.

2. Step 2: Verify leaf order correctness

   Visiting left subtree before right ensures leaves are collected left to right.

3. Step 3: Check for bugs

   No bug found; code correctly collects leaves in left to right order.

4. Final Answer:

   It visits leaves in left to right order correctly -> Option A

5. Quick Check:

   Leaf nodes collected left to right = It visits leaves in left to right order correctly [OK]

Hint: Left subtree first ensures left-to-right leaf order [OK]

Common Mistakes:

• Thinking recursion order is reversed
• Assuming non-leaf nodes are pushed
• Missing base case check

5. You want to perform boundary traversal on a binary tree but exclude all leaf nodes that have even values. Which approach correctly adapts the traversal?

hard

A. Modify leaf collection to skip leaves with even values during traversal

B. Skip left and right boundary nodes if they are even

C. Only collect leaves and ignore boundaries

D. Collect all nodes and filter even leaves after traversal

Solution

1. Step 1: Understand boundary traversal parts

   Boundary traversal includes left boundary, leaves, and right boundary.

2. Step 2: Identify where to exclude even leaves

   Only leaf nodes need filtering; boundaries remain unchanged.

3. Step 3: Adapt leaf collection function

   Modify leaf collection to check leaf value and skip if even.

4. Final Answer:

   Modify leaf collection to skip leaves with even values during traversal -> Option A

5. Quick Check:

   Filter leaves during traversal, not boundaries [OK]

Hint: Filter leaves during leaf collection step only [OK]

Common Mistakes:

• Filtering boundary nodes incorrectly
• Filtering after traversal causing extra work
• Ignoring leaf filtering altogether