Boundary Traversal of Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the Node structure for a binary tree.

DSA C++

struct Node {
    int data;
    Node* [1];
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Drag options to blanks, or click blank then click option'

Aleft

Bparent

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'next' instead of 'left' for the left child pointer.

The left child pointer in a binary tree node is commonly named 'left'.

2fill in blank

medium

Complete the code to check if a node is a leaf node.

DSA C++

bool isLeaf(Node* node) {
    return node != nullptr && node->left == nullptr && node->[1] == nullptr;
}

Drag options to blanks, or click blank then click option'

Aparent

Bright

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Checking wrong pointer names like 'parent' or 'next'.

A leaf node has no left or right children, so both pointers should be nullptr.

3fill in blank

hard

Fix the error in the function to add left boundary nodes (excluding leaves).

DSA C++

void addLeftBoundary(Node* root, vector<int>& res) {
    Node* cur = root->left;
    while (cur != nullptr) {
        if (!isLeaf(cur)) res.push_back(cur->[1]);
        if (cur->left != nullptr) cur = cur->left;
        else cur = cur->right;
    }
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cdata

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'right' instead of 'data' to get the node's value.

To add the node's value, we access the 'data' member, not the child pointers.

4fill in blank

hard

Fill both blanks to add right boundary nodes in reverse order (excluding leaves).

DSA C++

void addRightBoundary(Node* root, vector<int>& res) {
    Node* cur = root->[1];
    vector<int> tmp;
    while (cur != nullptr) {
        if (!isLeaf(cur)) tmp.push_back(cur->[2]);
        if (cur->right != nullptr) cur = cur->right;
        else cur = cur->left;
    }
    for (int i = tmp.size() - 1; i >= 0; --i) res.push_back(tmp[i]);
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cdata

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from left child or adding pointers instead of values.

Start from the right child of root and add node values (data) to tmp vector.

5fill in blank

hard

Fill in the blank to complete the boundary traversal function.

DSA C++

vector<int> boundaryTraversal(Node* root) {
    vector<int> res;
    if (root == nullptr) return res;
    if (!isLeaf(root)) res.push_back(root->[1]);
    addLeftBoundary(root, res);
    addLeaves(root, res);
    addRightBoundary(root, res);
    return res;
}

Drag options to blanks, or click blank then click option'

Adata

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Adding root pointer or child pointers instead of root's value.

The root's value is stored in 'data' and should be added if root is not a leaf.

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