Left Side View of Binary Tree in DSA C++ - Time & Space Complexity

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to know how the time needed to find the left side view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void leftSideView(TreeNode* root, vector<int>& result) {
    if (!root) return;
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i < size; i++) {
            TreeNode* node = q.front(); q.pop();
            if (i == 0) result.push_back(node->val);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
    }
}
    

This code visits each node level by level and records the first node of each level to get the left side view.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in a breadth-first manner.
• How many times: Exactly once per node, since each node is enqueued and dequeued once.

How Execution Grows With Input

As the number of nodes (n) increases, the code visits each node once, so the work grows directly with n.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows linearly as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the left side view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The code only visits the left children, so it runs faster than visiting all nodes."

[OK] Correct: The code actually visits every node once, including right children, because it needs to check all nodes to find the leftmost at each level.

Interview Connect

Understanding this linear time complexity helps you explain how tree traversals work efficiently and shows you can analyze level-order algorithms clearly.

Self-Check

"What if we used a depth-first search instead of breadth-first search? How would the time complexity change?"

Practice

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

easy

A. Nodes at the deepest level of the tree

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

C. All leaf nodes of the tree

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

Solution

1. Step 1: Understand the view perspective

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

2. Step 2: Identify visible nodes

   Only the first node encountered at each level from the left is visible in this view.

3. Final Answer:

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

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 node structure for a binary tree used in left side view traversal?

easy

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

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

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

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

Solution

1. Step 1: Check pointer usage for child nodes

   Child nodes must be pointers to allow dynamic linking in a tree.

2. Step 2: Validate data types

   Data is int, left and right children are pointers to Node, matching correct binary tree node structure.

3. Final Answer:

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

4. Quick Check:

   Use pointers for child nodes in tree struct [OK]

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

Common Mistakes:

• Using Node instead of Node* for children
• Using int type for child nodes
• Omitting pointers causing infinite size error

3. Consider the following binary tree:

      1
     / \
    2   3
   / \   \
  4   5   6

What is the left side view output of this tree?

medium

A. [1, 3, 6]

B. [1, 2, 4]

C. [1, 2, 5]

D. [1, 2, 6]

Solution

1. Step 1: Identify nodes visible from left at each level

   Level 1: 1, Level 2: 2 (left child), Level 3: 4 (left child of 2)

2. Step 2: Collect nodes in order

   Left side view nodes are [1, 2, 4]

3. Final Answer:

   [1, 2, 4] -> Option B

4. Quick Check:

   Left side view nodes = [1, 2, 4] [OK]

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Mixing nodes from different levels

4. The following C++ function is intended to print the left side view of a binary tree. What is the error?

void leftViewUtil(Node* root, int level, int &maxLevel) {
    if (!root) return;
    if (level > maxLevel) {
        cout << 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. The order of recursive calls should be right child first, then left child

B. maxLevel should be initialized to -1 instead of 0

C. No error; function correctly prints left side view

D. The function should print after both recursive calls

Solution

1. Step 1: Analyze traversal order and maxLevel usage

   The function visits left child first, then right child, printing first node at each level.

2. Step 2: Check initialization and printing logic

   maxLevel initialized to 0 and level starts at 1, so first level prints correctly; printing before recursion is correct.

3. Final Answer:

   No error; function correctly prints left side view -> Option C

4. Quick Check:

   Correct preorder traversal prints left view [OK]

Hint: Left child first recursion prints left view correctly [OK]

Common Mistakes:

• Changing recursion order breaks left view logic
• Incorrect maxLevel initialization causing missed nodes
• Printing after recursion misses first nodes

5. Given a binary tree where some levels have missing left children but have right children, which approach ensures correct left side view output?

hard

A. Traverse level order and pick first node at each level

B. Traverse preorder and print nodes only if left child exists

C. Traverse inorder and print nodes at odd levels

D. Traverse postorder and print leaf nodes only

Solution

1. Step 1: Understand problem with missing left children

   Left side view must include visible nodes even if left child is missing but right child exists.

2. Step 2: Choose traversal that captures first visible node per level

   Level order traversal visits nodes level by level, so picking first node at each level ensures correct left side view.

3. Final Answer:

   Traverse level order and pick first node at each level -> Option A

4. Quick Check:

   Level order first node per level = left side view [OK]

Hint: Use level order traversal, pick first node at each level [OK]

Common Mistakes:

• Ignoring right child when left child is missing
• Using preorder which may miss visible nodes
• Confusing inorder or postorder with left view logic