Boundary Traversal of Binary Tree in DSA Java - 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 grows as the tree gets bigger.

Specifically, how many steps does it take to visit all boundary nodes in a binary tree?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void boundaryTraversal(Node root) {
    if (root == null) return;
    System.out.print(root.data + " ");
    printLeftBoundary(root.left);
    printLeaves(root.left);
    printLeaves(root.right);
    printRightBoundary(root.right);
}

private void printLeftBoundary(Node node) {
    while (node != null) {
        if (!isLeaf(node)) System.out.print(node.data + " ");
        node = (node.left != null) ? node.left : node.right;
    }
}

private boolean isLeaf(Node node) {
    return (node.left == null && node.right == null);
}

This code prints the boundary nodes of a binary tree: left edge, leaves, and right edge.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the number of nodes (n) grows, the code visits each node once or twice but never more.

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: "Boundary traversal takes more than linear time because it does multiple traversals."

[OK] Correct: Even though it looks like multiple steps, each node is visited only a few times, so total work is still proportional to n.

Interview Connect

Understanding this helps you explain how tree traversals work efficiently and shows you can analyze combined traversals clearly.

Self-Check

"What if we used recursion instead of loops for left and right boundary? How would the time complexity change?"

Practice

1. What is the correct order of nodes visited in a boundary traversal of a binary tree?

easy

A. Root, leaves (left to right), left boundary (top-down), right boundary (bottom-up)

B. Leaves (left to right), root, left boundary (top-down), right boundary (bottom-up)

C. Root, left boundary (top-down), leaves (left to right), right boundary (bottom-up)

D. Left boundary (top-down), root, right boundary (bottom-up), leaves (left to right)

Solution

1. Step 1: Understand boundary traversal parts

   Boundary traversal visits root, then left boundary top-down, then leaves left to right, then right boundary bottom-up.

2. Step 2: Arrange traversal order

   Root is first, left boundary next, leaves after that, and right boundary last in reverse order.

3. Final Answer:

   Root, left boundary (top-down), leaves (left to right), right boundary (bottom-up) -> Option C

4. Quick Check:

   Boundary traversal order = Root + Left + Leaves + Right [OK]

Hint: Remember: Root, Left boundary, Leaves, Right boundary [OK]

Common Mistakes:

• Mixing order of leaves and boundaries
• Visiting right boundary top-down instead of bottom-up
• Skipping root node

2. Which of the following Java method signatures is correct for a function performing boundary traversal on a binary tree node?

easy

A. public void boundaryTraversal() Node root

B. void boundaryTraversal(Node root) public

C. public boundaryTraversal(Node root) void

D. public void boundaryTraversal(Node root)

Solution

1. Step 1: Recall Java method syntax

   Java methods start with access modifier, then return type, then method name and parameters.

2. Step 2: Match correct syntax

   public void boundaryTraversal(Node root) matches: public void boundaryTraversal(Node root). Others have wrong order or missing parts.

3. Final Answer:

   public void boundaryTraversal(Node root) -> Option D

4. Quick Check:

   Access modifier + return type + method name + parameters [OK]

Hint: Java method = access modifier + return type + name + params [OK]

Common Mistakes:

• Placing return type after method name
• Missing access modifier
• Incorrect parameter placement

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

        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 boundary nodes

   Root: 1; Left boundary top-down: 2 (excluding leaf 4); Leaves left to right: 4,7,8,9; Right boundary bottom-up: 6,3 (excluding leaves).

2. Step 2: Combine nodes without duplicates

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

3. Final Answer:

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

4. Quick Check:

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

Hint: Leaves included once; right boundary reversed; avoid duplicates [OK]

Common Mistakes:

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

4. Consider this Java method snippet for boundary traversal:

void printLeftBoundary(Node node) {
  if (node == null) return;
  System.out.print(node.data + " ");
  if (node.left != null) {
    printLeftBoundary(node.left);
  } else if (node.right != null) {
    printLeftBoundary(node.right);
  }
}

What is the bug in this code?

medium

A. It prints leaf nodes which should be excluded in left boundary

B. It does not print the root node

C. It causes infinite recursion on leaf nodes

D. It prints right boundary nodes instead of left

Solution

1. Step 1: Analyze leaf node printing

   The code prints node data before checking for children, thus printing leaf nodes as well.

2. Step 2: Identify leaf node exclusion missing

   Leaf nodes should not be printed in left boundary function, but this code prints them unconditionally after null check.

3. Final Answer:

   It prints leaf nodes which should be excluded in left boundary -> Option A

4. Quick Check:

   Left boundary excludes leaves, code prints them [OK]

Hint: Exclude leaf nodes in boundary functions to avoid duplicates [OK]

Common Mistakes:

• Not checking if node is leaf before printing
• Confusing left and right boundary logic
• Missing base case for null nodes

5. You have a binary tree where some nodes have only right children, and leaves are only on the rightmost path. How should you modify the boundary traversal algorithm to correctly print the boundary without duplicates?

hard

A. Skip printing left boundary if no left child; print leaves only once; print right boundary bottom-up excluding leaves

B. Print all nodes in preorder to cover all boundaries and leaves

C. Print left boundary top-down including leaves; skip right boundary

D. Print leaves first, then left and right boundaries in any order

Solution

1. Step 1: Understand tree structure

   Since leaves are only on rightmost path and some nodes have only right children, left boundary may be missing or partial.

2. Step 2: Modify traversal rules

   Skip left boundary if no left child; print leaves once; print right boundary bottom-up excluding leaves to avoid duplicates.

3. Step 3: Combine for correct boundary

   This ensures no duplicates and correct order for such skewed trees.

4. Final Answer:

   Skip printing left boundary if no left child; print leaves only once; print right boundary bottom-up excluding leaves -> Option A

5. Quick Check:

   Handle missing left boundary and leaf duplicates carefully [OK]

Hint: Skip missing boundaries; print leaves once; reverse right boundary [OK]

Common Mistakes:

• Printing leaves multiple times
• Ignoring missing left boundary
• Printing right boundary top-down causing wrong order