Boundary Traversal of Binary Tree in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a binary tree node struct.

DSA C

struct Node {
    int data;
    struct Node* [1];
    struct Node* right;
};

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

Achild

Bnext

Cprev

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'next' or 'prev' which are used in linked lists.

Using 'child' which is too generic.

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

2fill in blank

medium

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

DSA C

int isLeaf(struct Node* node) {
    return (node->left == NULL && node->[1] == NULL);
}

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

Anext

Bparent

Cright

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Checking 'next' or 'parent' pointers which do not exist in binary trees.

Checking only one child pointer.

A leaf node has both left and right pointers as NULL.

3fill in blank

hard

Fix the error in the function to print the left boundary excluding leaf nodes.

DSA C

void printLeftBoundary(struct Node* node) {
    if (node) {
        if (node->left) {
            printf("%d ", node->data);
            printLeftBoundary(node->[1]);
        } else if (node->right) {
            printf("%d ", node->data);
            printLeftBoundary(node->right);
        }
    }
}

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

Aright

Bleft

Cnext

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using right child pointer first which prints the wrong boundary.

Using non-existent pointers like 'next' or 'parent'.

To traverse the left boundary, we move down the left child first.

4fill in blank

hard

Fill both blanks to print the right boundary excluding leaf nodes in bottom-up order.

DSA C

void printRightBoundary(struct Node* node) {
    if (node) {
        if (node->[1]) {
            printRightBoundary(node->[1]);
            printf("%d ", node->data);
        } else if (node->[2]) {
            printRightBoundary(node->[2]);
            printf("%d ", node->data);
        }
    }
}

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

Aright

Bleft

Cnext

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing nodes before recursive call causing top-down order.

Using wrong child pointers.

For right boundary, traverse down the right child first, then left child if right is NULL.

5fill in blank

hard

Fill all three blanks to complete the boundary traversal function.

DSA C

void printBoundary(struct Node* root) {
    if (root) {
        printf("%d ", root->data);
        printLeftBoundary(root->[1]);
        printLeaves(root->[2]);
        printLeaves(root->[3]);
        printRightBoundary(root->right);
    }
}

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

Aleft

Bright

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'next' pointer which does not exist.

Mixing left and right pointers in wrong order.

Left boundary starts from root->left, leaves from both left and right subtrees, right boundary from root->right.

Practice

1. What is the main purpose of the Boundary Traversal of Binary Tree?

easy

A. To print all nodes on the boundary of the tree in an anti-clockwise direction

B. To print all nodes in level order

C. To print all leaf nodes only

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: left boundary, leaves, and right boundary.

2. Step 2: Compare with other traversals

   Level order prints all nodes level-wise, leaf traversal prints only leaves, in-order prints nodes in left-root-right order.

3. Final Answer:

   To print all nodes on the boundary of the tree in an anti-clockwise direction -> Option A

4. Quick Check:

   Boundary traversal = print edges anti-clockwise [OK]

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

Common Mistakes:

• Confusing boundary traversal with level order traversal
• Thinking it prints only leaf nodes
• Mixing it up with in-order traversal

2. Which of the following is the correct C function signature to perform boundary traversal on a binary tree node pointer named root?

easy

A. void boundaryTraversal(struct Node* root);

B. int boundaryTraversal(Node root);

C. void boundaryTraversal(Node root);

D. int boundaryTraversal(struct Node* root);

Solution

1. Step 1: Check parameter type for binary tree node

   Binary tree nodes are usually passed as pointers to struct Node in C.

2. Step 2: Check return type for traversal functions

   Traversal functions usually return void as they print nodes or store results externally.

3. Final Answer:

   void boundaryTraversal(struct Node* root); -> Option A

4. Quick Check:

   Pointer parameter + void return = void boundaryTraversal(struct Node* root); [OK]

Hint: Traversal functions usually take pointer and return void [OK]

Common Mistakes:

• Using struct keyword incorrectly
• Passing node by value instead of pointer
• Returning int instead of void

3. Consider the following binary tree:

        1
       / \
      2   3
     / \   \
    4   5   6
       / \  / \
      7  8 9  10

What is the output of the boundary traversal (anti-clockwise) of this tree?

medium

A. [1, 2, 4, 5, 7, 8, 9, 10, 6, 3]

B. [1, 2, 4, 7, 8, 9, 10, 3, 6]

C. [1, 2, 4, 7, 8, 9, 10, 6, 3]

D. [1, 2, 4, 7, 8, 5, 9, 10, 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, 10

3. Step 3: Identify right boundary nodes bottom-up excluding leaves

   Right boundary: 6, 3

4. Step 4: Combine in anti-clockwise order

   Boundary traversal: 1, 2, 4, 7, 8, 9, 10, 6, 3

5. Final Answer:

   [1, 2, 4, 7, 8, 9, 10, 6, 3] -> Option C

6. Quick Check:

   Left boundary + leaves + right boundary = [1, 2, 4, 7, 8, 9, 10, 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 C code snippet is intended to print the boundary traversal of a binary tree. What is the error?

void printLeftBoundary(Node* root) {
    if (root == NULL) return;
    printf("%d ", root->data);
    if (root->left) {
        printLeftBoundary(root->left);
    } else if (root->right) {
        printLeftBoundary(root->right);
    }
}

medium

A. It does not print the root node

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

C. It does not handle NULL root correctly

D. It prints nodes in bottom-up order instead of top-down

Solution

1. Step 1: Analyze printing condition in left boundary

   Function prints node data before recursive call, so top-down order is correct.

2. Step 2: Check if leaf nodes are excluded

   The printf is before the child checks, so leaf nodes are printed before returning.

3. Step 3: Identify error

   Leaf nodes should not be printed in left boundary function; they are printed separately in leaf traversal.

4. Final Answer:

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

5. Quick Check:

   Left boundary excludes leaves [OK]

Hint: Exclude leaf nodes in left boundary function [OK]

Common Mistakes:

• Printing leaf nodes in left or right boundary functions
• Not checking for NULL root
• Printing nodes in wrong order

5. Given a binary tree where some nodes have only one child, how should the boundary traversal be adapted to correctly include all boundary nodes without duplicates? Choose the best approach.

hard

A. Traverse all nodes in level order and print only those with one child

B. Traverse left and right boundaries including leaves, then print all leaves

C. Print all nodes in in-order traversal and remove duplicates afterwards

D. Traverse left boundary top-down excluding leaves, then all leaves left to right, then right boundary bottom-up excluding leaves, skipping duplicates

Solution

1. Step 1: Understand boundary traversal with single-child nodes

   Nodes with one child may appear in left or right boundary or as leaves.

2. Step 2: Avoid duplicates by excluding leaves in boundary traversals

   Left and right boundary traversals exclude leaves to avoid duplicates; leaves are printed separately.

3. Step 3: Combine traversals carefully

   Traverse left boundary top-down excluding leaves, then leaves left to right, then right boundary bottom-up excluding leaves, skipping duplicates.

4. Final Answer:

   Traverse left boundary top-down excluding leaves, then all leaves left to right, then right boundary bottom-up excluding leaves, skipping duplicates -> Option D

5. Quick Check:

   Exclude leaves in boundaries, print leaves separately [OK]

Hint: Exclude leaves in boundaries, print leaves separately [OK]

Common Mistakes:

• Including leaves in both boundary and leaf traversals causing duplicates
• Ignoring nodes with single child in boundary
• Printing nodes in wrong order