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?"