Boundary Traversal of Binary Tree in DSA Typescript - 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.

The question is: how does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function boundaryTraversal(root: TreeNode | null): number[] {
  if (!root) return [];
  const result: number[] = [];
  function leftBoundary(node: TreeNode | null) {
    if (!node || (!node.left && !node.right)) return;
    result.push(node.val);
    if (node.left) leftBoundary(node.left);
    else leftBoundary(node.right);
  }
  function leaves(node: TreeNode | null) {
    if (!node) return;
    leaves(node.left);
    if (!node.left && !node.right) result.push(node.val);
    leaves(node.right);
  }
  function rightBoundary(node: TreeNode | null) {
    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 to collect 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 in the tree increases, the code visits each node at most 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 number of steps grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to do the boundary traversal grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The boundary traversal only visits a few nodes, so it runs in constant time."

[OK] Correct: Even though only boundary nodes are collected, the code must check many nodes to find leaves and boundaries, so it still visits all nodes.

Interview Connect

Understanding this traversal helps you reason about tree algorithms and recursion, skills that are useful in many coding challenges and real projects.

Self-Check

"What if we changed the code to use an iterative approach with a stack 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 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: Identify traversal direction

   The traversal is done in anti-clockwise direction starting from root, left boundary, leaves, then right boundary.

3. Final Answer:

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

4. Quick Check:

   Boundary traversal = anti-clockwise boundary nodes [OK]

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

Common Mistakes:

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

2. Which of the following TypeScript function signatures correctly represents a function to perform boundary traversal of a binary tree?

easy

A. function boundaryTraversal(): number[]

B. function boundaryTraversal(root: TreeNode): void

C. function boundaryTraversal(root: TreeNode, result: number[]): void

D. function boundaryTraversal(root: TreeNode): number[]

Solution

1. Step 1: Analyze function purpose

   The function should take the root node and return an array of numbers representing boundary nodes.

2. Step 2: Check signatures

   function boundaryTraversal(root: TreeNode): number[] takes root and returns number array, matching expected usage. function boundaryTraversal(root: TreeNode): void returns void, so no output. function boundaryTraversal(root: TreeNode, result: number[]): void requires external array input, less common. function boundaryTraversal(): number[] has no input, so invalid.

3. Final Answer:

   function boundaryTraversal(root: TreeNode): number[] -> Option D

4. Quick Check:

   Function returns boundary nodes array [OK]

Hint: Function should return array of boundary nodes [OK]

Common Mistakes:

• Using void return type without output
• Missing root parameter
• Expecting external array input unnecessarily

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, 5, 7, 8, 9, 6, 3]

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

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

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 to top excluding leaves

   Right boundary: 3

4. Step 4: Combine in anti-clockwise order

   Boundary = left boundary + leaves (excluding duplicates) + right boundary = [1, 2, 4, 7, 8, 9, 6, 3]

5. Final Answer:

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

6. Quick Check:

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

Hint: Combine left boundary, leaves, right boundary anti-clockwise [OK]

Common Mistakes:

• Including leaf nodes twice
• Wrong order of right boundary nodes
• Missing root or boundary nodes

4. Consider the following TypeScript code snippet for boundary traversal. What is the bug?

function printBoundary(root: TreeNode | null) {
  if (!root) return;
  console.log(root.val);
  printLeftBoundary(root.left);
  printLeaves(root.left);
  printLeaves(root.right);
  printRightBoundary(root.right);
}

function printLeftBoundary(node: TreeNode | null) {
  if (!node || (!node.left && !node.right)) return;
  console.log(node.val);
  if (node.left) printLeftBoundary(node.left);
  else printLeftBoundary(node.right);
}

function printRightBoundary(node: TreeNode | null) {
  if (!node || (!node.left && !node.right)) return;
  if (node.right) printRightBoundary(node.right);
  else printRightBoundary(node.left);
  console.log(node.val);
}

medium

A. Left boundary includes leaf nodes which should be excluded

B. Root node is printed twice

C. Right boundary nodes are printed before recursive calls, causing wrong order

D. Leaves are printed twice because printLeaves is called on both left and right subtrees

Solution

1. Step 1: Analyze left boundary function

   printLeftBoundary prints node.val even if node is leaf, but leaves should be excluded from left boundary.

2. Step 2: Check leaf exclusion condition

   Condition (!node.left && !node.right) returns early, so leaves are excluded from printing in left boundary, so no bug here.

3. Step 3: Re-examine leaf printing calls

   printLeaves is called separately on left and right subtrees, so leaves are printed once only.

4. Step 4: Check right boundary printing order

   Right boundary prints after recursive calls, which is correct for bottom-up order.

5. Step 5: Identify actual bug

   Left boundary function prints node.val before recursive call, so leaf nodes on left boundary get printed twice: once in left boundary and once in leaves.

6. Final Answer:

   Left boundary includes leaf nodes which should be excluded -> Option A

7. Quick Check:

   Leaf nodes must not print in left boundary [OK]

Hint: Exclude leaves from left/right boundary printing [OK]

Common Mistakes:

• Printing leaves twice
• Printing right boundary in wrong order
• Printing root twice

5. You want to modify boundary traversal to print nodes in clockwise order starting from root. Which change is correct?

hard

A. Print root, then right boundary top-down, then leaves right to left, then left boundary bottom-up

B. Print root, then left boundary top-down, then leaves left to right, then right boundary bottom-up

C. Print root, then leaves left to right, then left boundary top-down, then right boundary bottom-up

D. Print root, then right boundary bottom-up, then leaves right to left, then left boundary top-down

Solution

1. Step 1: Understand clockwise boundary traversal order

   Clockwise traversal starts at root, then right boundary top-down, leaves right to left, then left boundary bottom-up.

2. Step 2: Match options with traversal order

   Print root, then right boundary top-down, then leaves right to left, then left boundary bottom-up matches this order exactly. Print root, then left boundary top-down, then leaves left to right, then right boundary bottom-up is anti-clockwise order. Options C and D mix orders incorrectly.

3. Final Answer:

   Print root, then right boundary top-down, then leaves right to left, then left boundary bottom-up -> Option A

4. Quick Check:

   Clockwise = root + right boundary + leaves + left boundary [OK]

Hint: Clockwise = root, right boundary, leaves, left boundary [OK]

Common Mistakes:

• Confusing clockwise with anti-clockwise order
• Printing boundaries in wrong direction
• Mixing leaf traversal order