Boundary Traversal of Binary Tree in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Boundary Traversal on a Simple Binary Tree

What is the output of the boundary traversal for the given binary tree?

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function boundaryTraversal(root) {
  if (!root) return [];
  const result = [];

  function isLeaf(node) {
    return !node.left && !node.right;
  }

  function addLeftBoundary(node) {
    let curr = node.left;
    while (curr) {
      if (!isLeaf(curr)) result.push(curr.data);
      if (curr.left) curr = curr.left;
      else curr = curr.right;
    }
  }

  function addLeaves(node) {
    if (isLeaf(node)) {
      result.push(node.data);
      return;
    }
    if (node.left) addLeaves(node.left);
    if (node.right) addLeaves(node.right);
  }

  function addRightBoundary(node) {
    let curr = node.right;
    const temp = [];
    while (curr) {
      if (!isLeaf(curr)) temp.push(curr.data);
      if (curr.right) curr = curr.right;
      else curr = curr.left;
    }
    for (let i = temp.length - 1; i >= 0; i--) {
      result.push(temp[i]);
    }
  }

  if (!isLeaf(root)) result.push(root.data);
  addLeftBoundary(root);
  addLeaves(root);
  addRightBoundary(root);

  return result;
}

// Tree construction
const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);

console.log(boundaryTraversal(root));

A [1, 2, 4, 5, 6, 7, 3]

B [1, 2, 4, 5, 7, 6, 3]

C [1, 2, 4, 5, 6, 3, 7]

D [1, 4, 5, 6, 7, 2, 3]

💡 Hint

Remember to include left boundary (excluding leaves), all leaves, then right boundary in reverse order.

The boundary traversal includes the root (1), left boundary nodes excluding leaves (2), all leaf nodes in left to right order (4, 5, 6, 7), and right boundary nodes excluding leaves in reverse order (3).

❓ Predict Output

intermediate

Boundary Traversal Output for Left Skewed Tree

What will be the output of the boundary traversal for this left skewed binary tree?

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function boundaryTraversal(root) {
  if (!root) return [];
  const result = [];

  function isLeaf(node) {
    return !node.left && !node.right;
  }

  function addLeftBoundary(node) {
    let curr = node.left;
    while (curr) {
      if (!isLeaf(curr)) result.push(curr.data);
      if (curr.left) curr = curr.left;
      else curr = curr.right;
    }
  }

  function addLeaves(node) {
    if (isLeaf(node)) {
      result.push(node.data);
      return;
    }
    if (node.left) addLeaves(node.left);
    if (node.right) addLeaves(node.right);
  }

  function addRightBoundary(node) {
    let curr = node.right;
    const temp = [];
    while (curr) {
      if (!isLeaf(curr)) temp.push(curr.data);
      if (curr.right) curr = curr.right;
      else curr = curr.left;
    }
    for (let i = temp.length - 1; i >= 0; i--) {
      result.push(temp[i]);
    }
  }

  if (!isLeaf(root)) result.push(root.data);
  addLeftBoundary(root);
  addLeaves(root);
  addRightBoundary(root);

  return result;
}

// Left skewed tree
const root = new Node(10);
root.left = new Node(20);
root.left.left = new Node(30);
root.left.left.left = new Node(40);

console.log(boundaryTraversal(root));

A [10, 40, 30, 20]

B [10, 20, 30]

C [40, 30, 20, 10]

D [10, 20, 30, 40]

💡 Hint

In a left skewed tree, left boundary and leaves overlap.

The root (10) is included first. The left boundary nodes excluding leaves are 20 and 30. The leaf node is 40. No right boundary nodes exist. So the traversal is [10, 20, 30, 40].

🔧 Debug

advanced

Identify the Bug in Boundary Traversal Code

The following code attempts to perform boundary traversal but produces incorrect output. What is the bug?

DSA Javascript

function addRightBoundary(node) {
  let curr = node.right;
  const temp = [];
  while (curr) {
    if (!isLeaf(curr)) temp.push(curr.data);
    if (curr.left) curr = curr.left;
    else curr = curr.right;
  }
  for (let i = temp.length - 1; i >= 0; i--) {
    result.push(temp[i]);
  }
}

A The function should start from node.left instead of node.right.

B The traversal of right boundary should prioritize right child first, not left child.

C The temp array should be reversed before pushing to result, but it is not reversed.

D The function should add leaves inside the loop, not outside.

💡 Hint

Check the order of child traversal in right boundary collection.

Right boundary traversal should move down the right child first to get the correct boundary nodes. The code incorrectly moves down the left child first, causing wrong nodes to be collected.

🧠 Conceptual

advanced

Understanding Leaf Node Identification in Boundary Traversal

In boundary traversal, why is it important to check if a node is a leaf before adding it to the left or right boundary?

A To avoid adding leaf nodes twice since leaves are added separately in the traversal.

B Because leaf nodes do not belong to the boundary of the tree.

C To ensure leaf nodes are only added in the right boundary traversal.

D To skip leaf nodes entirely from the traversal.

💡 Hint

Think about how leaves are handled separately in the traversal.

Leaf nodes are added once during the leaf collection step. Checking if a node is leaf before adding it to left or right boundary prevents duplication.

❓ Predict Output

expert

Boundary Traversal Output for Complex Tree with Missing Children

What is the output of the boundary traversal for the following binary tree?

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function boundaryTraversal(root) {
  if (!root) return [];
  const result = [];

  function isLeaf(node) {
    return !node.left && !node.right;
  }

  function addLeftBoundary(node) {
    let curr = node.left;
    while (curr) {
      if (!isLeaf(curr)) result.push(curr.data);
      if (curr.left) curr = curr.left;
      else curr = curr.right;
    }
  }

  function addLeaves(node) {
    if (isLeaf(node)) {
      result.push(node.data);
      return;
    }
    if (node.left) addLeaves(node.left);
    if (node.right) addLeaves(node.right);
  }

  function addRightBoundary(node) {
    let curr = node.right;
    const temp = [];
    while (curr) {
      if (!isLeaf(curr)) temp.push(curr.data);
      if (curr.right) curr = curr.right;
      else curr = curr.left;
    }
    for (let i = temp.length - 1; i >= 0; i--) {
      result.push(temp[i]);
    }
  }

  if (!isLeaf(root)) result.push(root.data);
  addLeftBoundary(root);
  addLeaves(root);
  addRightBoundary(root);

  return result;
}

// Complex tree
const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.left.right.left = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.right.left = new Node(8);

console.log(boundaryTraversal(root));

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

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

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

D [1, 2, 4, 5, 6, 7, 8, 3]

💡 Hint

Remember to add left boundary excluding leaves, then all leaves, then right boundary in reverse order.

The left boundary excluding leaves is [2, 4]. Leaves are [5, 6, 8]. Right boundary excluding leaves in reverse order is [7, 3]. Root is 1. So combined: [1, 2, 4, 5, 6, 8, 7, 3].

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