Why Left Side View of Binary Tree in DSA Go?

The Big Idea

Discover how to see the hidden branches of a tree from just one side without guessing!

The Scenario

Imagine you have a tall tree in your backyard. You want to see which branches you can spot if you stand on the left side of the tree. But the tree is big and has many branches overlapping each other.

Trying to list these branches by looking from the left side manually is like guessing which branch hides behind another.

The Problem

Manually checking each branch from the left side is slow and confusing. You might miss some branches or count the same branch twice. It's hard to keep track of which branch is visible and which is hidden behind others.

The Solution

The Left Side View of a Binary Tree helps you see exactly which nodes (branches) are visible when looking from the left side. It automatically picks the first node at each level, so you don't miss any visible branch and don't get confused by hidden ones.

Before vs After

✗ Before

func printLeftSideManual(root *Node) {
    // Manually guess visible nodes
    fmt.Println(root.Value)
    if root.Left != nil {
        fmt.Println(root.Left.Value)
    }
    if root.Left != nil && root.Left.Left != nil {
        fmt.Println(root.Left.Left.Value)
    }
}

✓ After

func printLeftSideView(root *Node) {
    // Use level order traversal
    // Print first node of each level
    // Automatically handles visibility
}

What It Enables

This concept lets you quickly find all nodes visible from the left side of a tree, making complex tree structures easy to understand and visualize.

Real Life Example

Think of a family tree where you want to see the oldest ancestor at each generation from the left side. The Left Side View shows you exactly those ancestors without confusion.

Key Takeaways

Manual checking of visible nodes is confusing and error-prone.

Left Side View automatically finds the first visible node at each level.

This helps visualize and understand tree structures clearly.

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