Left Side View of Binary Tree in DSA C++

Mental Model

We want to see the nodes visible when looking at the tree from the left side, which means the first node at each level.

Analogy: Imagine standing on the left side of a tall building with many floors and windows; you only see the first window on each floor from your side.

      1
     / \
    2   3
   / \   \
  4   5   6

Left side view: 1 -> 2 -> 4 -> null

Dry Run Walkthrough

Input: Binary tree: 1 / \ 2 3 / \ \ 4 5 6

Goal: Print the left side view nodes: the first node visible at each level from left side

Step 1: Start level order traversal at root (level 0)

Level 0: [1]
Left side node: 1

Why: Root is always visible from left side

Step 2: Move to level 1, nodes are 2 and 3

Level 1: [2, 3]
Left side node: 2

Why: At level 1, first node from left is 2

Step 3: Move to level 2, nodes are 4, 5, and 6

Level 2: [4, 5, 6]
Left side node: 4

Why: At level 2, first node from left is 4

Result:

Left side view: 1 -> 2 -> 4 -> null

Annotated Code

DSA C++

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

vector<int> leftSideView(TreeNode* root) {
    vector<int> result;
    if (!root) return result; // handle empty tree

    queue<TreeNode*> q;
    q.push(root);

    while (!q.empty()) {
        int levelSize = q.size();
        for (int i = 0; i < levelSize; i++) {
            TreeNode* node = q.front();
            q.pop();

            if (i == 0) {
                result.push_back(node->val); // first node at this level
            }

            if (node->left) q.push(node->left); // enqueue left child
            if (node->right) q.push(node->right); // enqueue right child
        }
    }
    return result;
}

int main() {
    // Construct the tree from dry_run input
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->right->right = new TreeNode(6);

    vector<int> view = leftSideView(root);
    for (int val : view) {
        cout << val << " -> ";
    }
    cout << "null" << endl;
    return 0;
}

if (!root) return result; // handle empty tree

return empty if tree is empty

q.push(root);

start BFS from root

int levelSize = q.size();

get number of nodes at current level

for (int i = 0; i < levelSize; i++) {

iterate over all nodes at this level

TreeNode* node = q.front(); q.pop();

process current node and remove from queue

if (i == 0) { result.push_back(node->val); }

record first node at this level for left view

if (node->left) q.push(node->left);

enqueue left child for next level

if (node->right) q.push(node->right);

enqueue right child for next level

Output

1 -> 2 -> 4 -> null

Complexity Analysis

Time: O(n) because we visit each node exactly once in the breadth-first traversal

Space: O(w) where w is the maximum width of the tree, due to the queue storing nodes at each level

vs Alternative: Compared to a depth-first approach with extra bookkeeping, BFS naturally processes nodes level by level, making it simpler and efficient for this problem

Edge Cases

Empty tree

Returns empty list since there are no nodes

DSA C++

if (!root) return result; // handle empty tree

Tree with only root node

Returns list with only root value

DSA C++

if (i == 0) { result.push_back(node->val); }

Tree with nodes only on right side

Still returns first node at each level, which will be the right nodes

DSA C++

if (i == 0) { result.push_back(node->val); }

When to Use This Pattern

When asked to find nodes visible from one side of a tree, use level order traversal and pick the first node at each level to get the left side view.

Common Mistakes

Mistake: Collecting all nodes at each level instead of only the first node
Fix: Only add the node value to result when processing the first node at each level (i == 0)

Mistake: Using depth-first traversal without tracking levels properly
Fix: Use breadth-first traversal (queue) to process nodes level by level

Summary

Prints the first node visible at each level when looking at the tree from the left side.

Use when you need to see the leftmost nodes of each level in a binary tree.

The key insight is to do a level order traversal and record only the first node at each level.

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