Boundary Traversal of Binary Tree in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Boundary Traversal of a binary tree?

Boundary Traversal means visiting all the nodes on the boundary of the tree in an anti-clockwise direction starting from the root. It includes the left boundary, leaves, and right boundary.

beginner

Which parts of the binary tree are included in the boundary traversal?

The boundary traversal includes three parts: 1) Left boundary (excluding leaves), 2) All leaf nodes (left to right), 3) Right boundary (excluding leaves, in bottom-up order).

intermediate

How do you avoid printing leaf nodes twice in boundary traversal?

Leaf nodes are printed only once during the leaf traversal step. Left and right boundary traversals exclude leaf nodes to avoid duplicates.

beginner

What is the order of traversal in boundary traversal of a binary tree?

The order is: root node → left boundary (top-down) → leaf nodes (left to right) → right boundary (bottom-up).

intermediate

Why is the right boundary printed in bottom-up order in boundary traversal?

The right boundary is printed bottom-up to maintain the anti-clockwise direction of traversal around the tree boundary.

Which nodes are NOT included in the left boundary traversal?

A All nodes on the left edge

B Root node

C Right child nodes

D Leaf nodes

Leaf nodes are excluded from the left boundary traversal to avoid duplication since they are printed separately.

In boundary traversal, when are leaf nodes printed?

A Before left boundary

B After left boundary and before right boundary

C After right boundary

D Only if they are on the left boundary

Leaf nodes are printed after the left boundary traversal and before the right boundary traversal.

What is the traversal direction of the right boundary in boundary traversal?

A Bottom-up

B Top-down

C Left to right

D Right to left

The right boundary is traversed bottom-up to maintain the anti-clockwise order.

Which node is always the first node printed in boundary traversal?

A Root node

B Rightmost leaf

C Leftmost leaf

D Right boundary node

The root node is always printed first in boundary traversal.

If a binary tree has only one node, what is the boundary traversal output?

A Empty

B Left boundary only

C Root node only

D Right boundary only

If there is only one node, it is the root and also a leaf, so it is printed once.

Explain the steps to perform boundary traversal of a binary tree.

Think about the three parts of the boundary and their order.

How do you handle leaf nodes during boundary traversal to avoid duplicates?

Consider where leaf nodes appear in the traversal.

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