Left Side View of Binary Tree in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Left Side View of a Binary Tree?

The Left Side View of a Binary Tree is the set of nodes visible when the tree is seen from the left side. It includes the leftmost node at each level of the tree.

beginner

Which tree traversal method is commonly used to find the Left Side View of a Binary Tree?

Level Order Traversal (Breadth-First Search) is commonly used. It processes nodes level by level and helps identify the first node at each level, which forms the left side view.

beginner

In Go, which data structure is typically used to implement the queue needed for level order traversal?

A slice (dynamic array) is typically used as a queue in Go. Elements are appended at the end and removed from the front to simulate queue behavior.

beginner

Why do we only take the first node at each level for the Left Side View?

Because the first node at each level is the leftmost node visible from the left side. Nodes after it are hidden behind this node when viewed from the left.

intermediate

What is the time complexity of finding the Left Side View of a Binary Tree using level order traversal?

The time complexity is O(n), where n is the number of nodes in the tree, because each node is visited exactly once.

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

A Nodes visible from the left side at each level

B Nodes visible from the right side at each level

C All leaf nodes of the tree

D Nodes with only left children

The left side view shows the nodes visible when looking at the tree from the left side, which are the leftmost nodes at each level.

Which traversal technique is best suited to find the Left Side View?

A Preorder Traversal

B Level Order Traversal

C Inorder Traversal

D Postorder Traversal

Level Order Traversal visits nodes level by level, making it easy to pick the first node at each level for the left side view.

In Go, what is a simple way to implement a queue for level order traversal?

A Using a slice with append and slicing

B Using a map

C Using a linked list package

D Using a stack

A slice can be used as a queue by appending elements at the end and removing from the front using slicing.

If a binary tree has 10 nodes, what is the time complexity to find its left side view?

A O(1)

B O(log n)

C O(n^2)

D O(n)

Each node is visited once during level order traversal, so the time complexity is O(n).

Why do we ignore nodes after the first node at each level when finding the left side view?

A They are always leaf nodes

B They are not part of the tree

C They are hidden behind the first node from the left view

D They are duplicates

Nodes after the first node at each level are hidden behind it when viewed from the left side.

Explain how to find the Left Side View of a Binary Tree using level order traversal.

Think about visiting nodes level-wise and picking the first node you see at each level.

Describe the role of a queue in implementing the Left Side View algorithm in Go.

Consider how you visit nodes level by level and keep track of them.

Practice

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

easy

A. All leaf nodes of the tree

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

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

D. Nodes at the deepest level only

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.

3. Final Answer:

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

4. Quick Check:

   Left side view = nodes seen from left [OK]

Hint: Think of standing on the left side and seeing nodes [OK]

Common Mistakes:

• Confusing left side view with right side view
• Including all leaf nodes instead of first nodes per level
• Thinking it shows deepest nodes only

2. Which Go code snippet correctly declares a binary tree node struct for use in left side view traversal?

easy

A. type TreeNode struct { Value int; LeftNode, RightNode *TreeNode }

B. type TreeNode struct { Val int; Left, Right *TreeNode }

C. struct TreeNode { int Val; TreeNode* Left; TreeNode* Right; }

D. class TreeNode { int val; TreeNode left; TreeNode right; }

Solution

1. Step 1: Identify Go struct syntax

   Go uses 'type Name struct { fields }' with field types after names.

2. Step 2: Check pointer usage for child nodes

   Child nodes are pointers to TreeNode, written as '*TreeNode'.

3. Final Answer:

   type TreeNode struct { Val int; Left, Right *TreeNode } -> Option B

4. Quick Check:

   Go struct with pointers = type TreeNode struct { Val int; Left, Right *TreeNode } [OK]

Hint: Go structs use 'type' and pointers with * for child nodes [OK]

Common Mistakes:

• Using C or Java syntax instead of Go
• Missing pointer * for child nodes
• Incorrect field names or order

3. What is the left side view output of this binary tree?

    1
   / \
  2   3
   \   \
    4   5

medium

A. [1, 2, 4]

B. [1, 3, 5]

C. [1, 2, 3]

D. [1, 4, 5]

Solution

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

   Level 1: Node 1; Level 2: Node 2 (left child); Level 3: Node 4 (right child of 2, visible from left).

2. Step 2: Collect nodes in order

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

3. Final Answer:

   [1, 2, 4] -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Confusing level order

4. Given this Go function to get left side view, what is the bug?

func leftSideView(root *TreeNode) []int {
    var result []int
    var dfs func(node *TreeNode, level int)
    dfs = func(node *TreeNode, level int) {
        if node == nil {
            return
        }
        if level == len(result) {
            result = append(result, node.Val)
        }
        dfs(node.Right, level+1)
        dfs(node.Left, level+1)
    }
    dfs(root, 0)
    return result
}

medium

A. Traverse right child before left, so left side nodes missed

B. Missing base case for nil node

C. Appending node.Val at wrong condition

D. Not initializing result slice

Solution

1. Step 1: Analyze traversal order

   The function visits right child before left child, so right nodes are added first.

2. Step 2: Understand left side view logic

   To get left side view, left child must be visited before right child to capture leftmost nodes first.

3. Final Answer:

   Traverse right child before left, so left side nodes missed -> Option A

4. Quick Check:

   Left side view needs left-first traversal [OK]

Hint: Visit left child before right for left side view [OK]

Common Mistakes:

• Swapping left and right traversal order
• Ignoring base case (though present here)
• Appending values incorrectly

5. You want to find the left side view of a binary tree but only have access to its level order traversal array with nulls for missing nodes:

[1, 2, 3, null, 4, null, 5]

Which approach correctly finds the left side view?

hard

A. Sort the array and pick first element of each value group

B. Traverse the array directly and pick all non-null nodes

C. Use DFS on array indices to find leftmost nodes

D. Rebuild the tree from level order, then do BFS picking first node at each level

Solution

1. Step 1: Understand input format

   Level order array with nulls represents tree nodes level by level, including missing nodes.

2. Step 2: Rebuild tree structure

   Reconstruct the binary tree from the array to restore parent-child links.

3. Step 3: Perform BFS for left side view

   Use breadth-first search and pick the first node at each level for left side view.

4. Final Answer:

   Rebuild the tree from level order, then do BFS picking first node at each level -> Option D

5. Quick Check:

   Rebuild + BFS first node per level = left side view [OK]

Hint: Rebuild tree first, then BFS for leftmost nodes per level [OK]

Common Mistakes:

• Trying to pick nodes directly from array without tree structure
• Using DFS on array indices which is invalid
• Sorting array which loses tree structure