Left Side View of Binary Tree in DSA Go

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; 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 with left child 2 and right child 3; 2 has children 4 and 5; 3 has right child 6

Goal: Find the leftmost node at each level to form the left side view

Step 1: Start at root node 1, add 1 to left view

Level 0: [1]
Left view: 1

Why: Root is always visible from the left side

Step 2: Go to level 1, nodes are 2 and 3; add 2 to left view

Level 1: [2, 3]
Left view: 1 -> 2

Why: At level 1, the leftmost node is 2

Step 3: Go to level 2, nodes are 4, 5, and 6; add 4 to left view

Level 2: [4, 5, 6]
Left view: 1 -> 2 -> 4

Why: At level 2, the leftmost node is 4

Result:

1 -> 2 -> 4 -> null

Annotated Code

DSA Go

package main

import (
	"fmt"
)

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

func leftSideView(root *TreeNode) []int {
	if root == nil {
		return []int{}
	}
	result := []int{}
	queue := []*TreeNode{root}

	for len(queue) > 0 {
		n := len(queue)
		for i := 0; i < n; i++ {
			curr := queue[0]
			queue = queue[1:]

			if i == 0 {
				result = append(result, curr.Val) // add first node of this level
			}

			if curr.Left != nil {
				queue = append(queue, curr.Left) // enqueue left child
			}
			if curr.Right != nil {
				queue = append(queue, curr.Right) // enqueue right child
			}
		}
	}
	return 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}
	root.Right.Right = &TreeNode{Val: 6}

	view := leftSideView(root)
	for _, v := range view {
		fmt.Printf("%d -> ", v)
	}
	fmt.Print("null\n")
}

if root == nil { return []int{} }

handle empty tree edge case

queue := []*TreeNode{root}

start BFS with root node in queue

for len(queue) > 0 { n := len(queue) }

process nodes level by level

curr := queue[0]; queue = queue[1:]

dequeue current node for processing

if i == 0 { result = append(result, curr.Val) }

add first node of each level to left view

if curr.Left != nil { queue = append(queue, curr.Left) }

enqueue left child for next level

if curr.Right != nil { queue = append(queue, curr.Right) }

enqueue right child for next level

Output

1 -> 2 -> 4 -> null

Complexity Analysis

Time: O(n) because we visit each node once during the breadth-first search

Space: O(n) because the queue can hold up to all nodes at the widest level

vs Alternative: Compared to a recursive DFS approach, BFS naturally processes nodes level by level, making it easier to pick the first node at each level without extra bookkeeping

Edge Cases

Empty tree (root is nil)

Returns an empty list since there are no nodes to view

DSA Go

if root == nil { return []int{} }

Tree with only one node

Returns a list with just that node's value

DSA Go

queue := []*TreeNode{root}

Tree where some nodes have only right children

Still correctly picks the leftmost node at each level, which may be a right child if no left child exists

DSA Go

if curr.Left != nil { queue = append(queue, curr.Left) }

When to Use This Pattern

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

Common Mistakes

Mistake: Adding all nodes at each level instead of only the first one
Fix: Add only the first node (i == 0) of each level to the result list

Mistake: Using preorder DFS without tracking levels properly
Fix: Use BFS or track levels carefully in DFS to ensure only the leftmost node per level is added

Summary

Finds the leftmost node at each level of a binary tree to show the left side view.

Use when you need to see which nodes are visible when looking at the tree from the left side.

The key insight is to traverse level by level and pick only the first node encountered at each level.

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