Left Side View of Binary Tree in DSA Go - Build from Scratch

Left Side View of Binary Tree

📖 Scenario: Imagine you have a family tree represented as a binary tree. You want to see only the leftmost family members at each generation level.

🎯 Goal: Build a program that finds the left side view of a binary tree. This means printing the nodes you see when looking at the tree from the left side.

📋 What You'll Learn

Create a binary tree with specific nodes

Use a variable to track the current level during traversal

Implement a function to collect the left side view nodes

Print the left side view nodes in order

💡 Why This Matters

🌍 Real World

Left side view of a binary tree helps visualize hierarchical data from a specific angle, useful in family trees, organizational charts, and UI layouts.

💼 Career

Understanding tree traversal and views is important for software engineers working with hierarchical data structures, UI rendering, and algorithms.

1

Create the binary tree structure and nodes

Define a struct called TreeNode with Val int, Left *TreeNode, and Right *TreeNode. Then create the binary tree with these exact nodes and connections:
root with value 1,
root.Left with value 2,
root.Right with value 3,
root.Left.Left with value 4,
root.Left.Right with value 5.

DSA Go

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

// Create the binary tree nodes and connect them
// Your code here

Hint

Start by defining the TreeNode struct. Then create the root node and assign its left and right children as shown.

2

Add a variable to track the current level

Inside main, create a variable called maxLevel and set it to 0. This will help track the deepest level visited so far.

DSA Go

package main

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}

    // Create maxLevel variable
    // Your code here
}

Hint

Declare maxLevel as an integer and initialize it to zero inside the main function.

3

Implement the function to get the left side view

Write a function called leftView that takes root *TreeNode, level int, maxLevel *int, and result *[]int. Inside it, if root is nil, return immediately. If level is greater than *maxLevel, append root.Val to *result and update *maxLevel = level. Then recursively call leftView for root.Left and root.Right with level+1.

DSA Go

package main

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func leftView(root *TreeNode, level int, maxLevel *int, result *[]int) {
    // Your code here
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}

    maxLevel := 0
}

Hint

Use recursion to visit nodes. Check if the current level is deeper than maxLevel. If yes, add the node's value to result and update maxLevel. Visit left child first, then right child.

4

Call the function and print the left side view

In main, create a slice called result of type []int. Call leftView(root, 1, &maxLevel, &result). Then print result using fmt.Println(result). Import "fmt" at the top.

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func leftView(root *TreeNode, level int, maxLevel *int, result *[]int) {
    if root == nil {
        return
    }
    if level > *maxLevel {
        *result = append(*result, root.Val)
        *maxLevel = level
    }
    leftView(root.Left, level+1, maxLevel, result)
    leftView(root.Right, level+1, maxLevel, result)
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Right = &TreeNode{Val: 5}

    maxLevel := 0

    // Create result slice and call leftView
    // Print the result
    // Your code here
}

Hint

Create an empty slice result. Call leftView starting at level 1. Print the result slice to see the left side view.

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