Left Side View of Binary Tree in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is NULL.

DSA C

if (root [1] NULL) {
    return;
}

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

A!=

B=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '==' causes assignment instead of comparison.

Using '!=' reverses the condition.

We use '==' to compare if the pointer root is NULL.

2fill in blank

medium

Complete the code to print the node's data.

DSA C

printf("%d ", root->[1]);

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

Aright

Bdata

Cleft

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'right' prints pointers, not values.

Using 'value' when the struct uses 'data'.

The node's value is stored in the data field.

3fill in blank

hard

Fix the error in the recursive call to traverse the left subtree.

DSA C

leftSideViewUtil(root->[1], level + 1, &maxLevel);

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

Aleft

Bnext

Cdata

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' causes traversal of the wrong subtree first.

Using 'data' or 'next' are invalid for tree traversal.

To get the left side view, we first traverse the left child using root->left.

4fill in blank

hard

Fill both blanks to update the maxLevel and check if current level is first visit.

DSA C

if (level > *[1]) {
    printf("%d ", root->data);
    *[2] = level;
}

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

AmaxLevel

Blevel

Croot

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for comparison and update causes logic errors.

Using 'level' instead of '*maxLevel' for comparison.

We compare level with *maxLevel and update *maxLevel when a new level is visited.

5fill in blank

hard

Fill all three blanks to complete the function signature and initial call.

DSA C

void leftSideViewUtil(struct Node* [1], int level, int* [2]) {
    if ([1] == NULL) return;
    if (level > *[2]) {
        printf("%d ", [1]->data);
        *[2] = level;
    }
    leftSideViewUtil([1]->left, level + 1, [2]);
    leftSideViewUtil([1]->right, level + 1, [2]);
}

void leftSideView(struct Node* root) {
    int maxLevel = 0;
    leftSideViewUtil(root, [3], &maxLevel);
}

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

Aroot

BmaxLevel

C1

Dlevel

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect variable names in the function signature.

Starting level at 0 instead of 1.

The function uses root as the node pointer, maxLevel as the pointer to max level, and starts traversal at level 1.

Practice

1. What does the Left Side View of a binary tree represent?

easy

A. Nodes visible when the tree is viewed from the left side

B. Nodes visible when the tree is viewed from the right side

C. All leaf nodes of the tree

D. Nodes at the deepest level of the tree

Solution

1. Step 1: Understand the view concept

   The left side view shows nodes visible when looking at the tree from the left side.

2. Step 2: Differentiate from other views

   Right side view shows nodes from the right, leaf nodes are only the bottom nodes, and deepest level nodes are not necessarily visible from left.

3. Final Answer:

   Nodes visible when the tree is viewed from the left side -> Option A

4. Quick Check:

   Left side view = nodes visible from left [OK]

Hint: Left side view shows first node at each level from left [OK]

Common Mistakes:

• Confusing left side view with right side view
• Thinking it shows all leaf nodes
• Assuming it shows deepest nodes only

2. Which of the following C code snippets correctly declares a binary tree node structure for computing the left side view?

easy

A. struct Node { int data; struct Node left; struct Node right; };

B. typedef struct Node { int data; struct Node* left; struct Node* right; } Node;

C. struct Node { int data; Node* left; Node* right; }; typedef Node;

D. typedef struct { int data; struct Node* left; struct Node* right; } Node;

Solution

1. Step 1: Check correct struct declaration

   typedef struct Node { int data; struct Node* left; struct Node* right; } Node; correctly declares a struct with pointers to left and right child nodes.

2. Step 2: Identify errors in other options

   struct Node { int data; Node* left; Node* right; }; typedef Node; misses struct keyword before Node pointers, C uses struct Node instead of pointers causing infinite size, D misses struct name for pointers.

3. Final Answer:

   typedef struct Node { int data; struct Node* left; struct Node* right; } Node; -> Option B

4. Quick Check:

   Correct struct with pointers = typedef struct Node { int data; struct Node* left; struct Node* right; } Node; [OK]

Hint: Use pointers for left and right child nodes in struct [OK]

Common Mistakes:

• Using struct members instead of pointers causing infinite size
• Missing struct keyword before pointer types
• Incorrect typedef syntax

3. Given the following binary tree:

      1
     / \
    2   3
   /     \
  4       5

What is the left side view output printed by this C code?

void leftViewUtil(Node* root, int level, int* maxLevel) {
  if (!root) return;
  if (level > *maxLevel) {
    printf("%d ", root->data);
    *maxLevel = level;
  }
  leftViewUtil(root->left, level + 1, maxLevel);
  leftViewUtil(root->right, level + 1, maxLevel);
}

void leftView(Node* root) {
  int maxLevel = 0;
  leftViewUtil(root, 1, &maxLevel);
}

medium

A. 1 3 5

B. 1 2 5

C. 1 2 4

D. 1 4 5

Solution

1. Step 1: Trace recursive calls and printing

   At level 1, print 1; level 2, print 2 (left child); level 3, print 4 (left child of 2). Right children at same levels are ignored since maxLevel is updated.

2. Step 2: Confirm output sequence

   Printed nodes are 1, 2, 4 in order with spaces.

3. Final Answer:

   1 2 4 -> Option C

4. Quick Check:

   Left side view nodes = 1 2 4 [OK]

Hint: Print first node at each level during preorder left-first traversal [OK]

Common Mistakes:

• Printing right child nodes before left
• Not updating maxLevel correctly
• Confusing left side view with right side view

4. Identify the bug in this C function intended to print the left side view of a binary tree:

void leftViewUtil(Node* root, int level, int* maxLevel) {
  if (root == NULL) return;
  if (level >= *maxLevel) {
    printf("%d ", root->data);
    *maxLevel = level;
  }
  leftViewUtil(root->left, level + 1, maxLevel);
  leftViewUtil(root->right, level + 1, maxLevel);
}

medium

A. The condition should be 'level > *maxLevel' instead of 'level >= *maxLevel'

B. The function should print after recursive calls, not before

C. The base case should check 'if (root != NULL)' instead of 'if (root == NULL)'

D. The maxLevel pointer should be passed by value, not by reference

Solution

1. Step 1: Analyze the condition for printing

   The condition 'level >= *maxLevel' causes printing nodes multiple times at the same level, which is incorrect.

2. Step 2: Correct condition usage

   It should be 'level > *maxLevel' to print only the first node encountered at each level.

3. Final Answer:

   The condition should be 'level > *maxLevel' instead of 'level >= *maxLevel' -> Option A

4. Quick Check:

   Use strict greater than for first node per level [OK]

Hint: Use '>' not '>=' to print only first node at each level [OK]

Common Mistakes:

• Using '>=' causes duplicate prints at same level
• Printing after recursion changes order
• Incorrect base case logic

5. You want to print the left side view of a binary tree using a queue for level order traversal. Which approach correctly achieves this?

1. Enqueue root
2. While queue not empty:
   a. Get number of nodes at current level (size)
   b. For i in 0 to size-1:
       - Dequeue node
       - If i == 0, print node data
       - Enqueue left child if exists
       - Enqueue right child if exists

What is the output for this tree?

      10
     /  \
    20   30
      \    \
      40    50

hard

A. 10 40 50

B. 10 30 50

C. 10 20 50

D. 10 20 40

Solution

1. Step 1: Perform level order traversal with printing first node per level

   Level 1: print 10; Level 2: print 20 (first node at level); Level 3: print 40 (first node at level).

2. Step 2: Confirm output sequence

   Nodes printed in order: 10 20 40 with spaces.

3. Final Answer:

   10 20 40 -> Option D

4. Quick Check:

   Left side view by level order = 10 20 40 [OK]

Hint: Print first node at each level during BFS traversal [OK]

Common Mistakes:

• Printing last node instead of first at each level
• Mixing left and right child enqueue order
• Not counting nodes per level correctly