Left Side View of Binary Tree in DSA Go - Execution Trace

Concept Flow - Left Side View of Binary Tree

Start at root node

↓

Initialize queue with root

↓

While queue not empty

↓

Get number of nodes at current level

↓

For each node in level

↓

Pop node

↓

Add left child to queue if exists

↓

Add right child to queue if exists

↓

Repeat for next level

↓

End when queue empty

↓

Return collected left view nodes

We start from the root and use a queue to traverse the tree level by level. For each level, we record the first node seen as part of the left side view.

Execution Sample

DSA Go

func leftSideView(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    queue := []*TreeNode{root}
    var result []int
    for len(queue) > 0 {
        levelSize := len(queue)
        for i := 0; i < levelSize; i++ {
            node := queue[0]
            queue = queue[1:]
            if i == 0 {
                result = append(result, node.Val)
            }
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
    }
    return result
}

This code collects the left side view nodes of a binary tree using level order traversal.

Execution Table

Step	Operation	Nodes in Queue	Pointer Changes	Visual State (Tree + Left View)
1	Initialize queue with root (1)	[1]	queue = [1]	Tree: 1 Left View: []
2	Start level 1, levelSize=1	[1]	levelSize=1	Tree: 1 Left View: []
3	Pop node 1 (first in level)	[]	queue=[]	Tree: 1 Left View: [1]
4	Add left child 2 to queue	[2]	queue=[2]	Tree: 1 Left View: [1]
5	Add right child 3 to queue	[2,3]	queue=[2,3]	Tree: 1 Left View: [1]
6	Start level 2, levelSize=2	[2,3]	levelSize=2	Tree: 1 Left View: [1]
7	Pop node 2 (first in level)	[3]	queue=[3]	Tree: 1 Left View: [1,2]
8	Add left child 4 to queue	[3,4]	queue=[3,4]	Tree: 1 Left View: [1,2]
9	Add right child 5 to queue	[3,4,5]	queue=[3,4,5]	Tree: 1 Left View: [1,2]
10	Pop node 3 (second in level)	[4,5]	queue=[4,5]	Tree: 1 Left View: [1,2]
11	No children to add	[4,5]	queue=[4,5]	Tree: 1 Left View: [1,2]
12	Start level 3, levelSize=2	[4,5]	levelSize=2	Tree: 1 Left View: [1,2]
13	Pop node 4 (first in level)	[5]	queue=[5]	Tree: 1 Left View: [1,2,4]
14	No children to add	[5]	queue=[5]	Tree: 1 Left View: [1,2,4]
15	Pop node 5 (second in level)	[]	queue=[]	Tree: 1 Left View: [1,2,4]
16	No children to add	[]	queue=[]	Tree: 1 Left View: [1,2,4]
17	Queue empty, end traversal	[]	queue=[]	Tree: 1 Left View: [1,2,4]

💡 Queue is empty, all levels processed, left view collected

Variable Tracker

Variable	Start	After Step 1	After Step 5	After Step 10	After Step 15	Final
queue	[]	[1]	[2,3]	[4,5]	[]	[]
levelSize	0	1	2	2	0	0
result (left view)	[]	[]	[1]	[1,2]	[1,2,4]	[1,2,4]

Key Moments - 3 Insights

Why do we add only the first node of each level to the left view?

Because the first node at each level is the leftmost node visible from the left side. This is shown in execution_table rows 3, 7, and 13 where the first node popped at each level is added to the left view.

Why do we use a queue to traverse the tree?

A queue helps us process nodes level by level (breadth-first). This ensures we see nodes in order from left to right at each level, as shown in the queue states in execution_table rows 1, 5, 10, and 15.

What happens if the tree is empty?

If the root is nil, the function returns an empty list immediately, so no traversal happens. This is the start condition before step 1.

Visual Quiz - 3 Questions

Look at the execution_table at step 7. What is the left view after popping node 2?

A [1]

B [1, 2]

C [1, 3]

D [2]

💡 Hint

Check the 'Visual State' column at step 7 in execution_table.

At which step does the queue become empty, ending the traversal?

A Step 17

B Step 15

C Step 10

D Step 5

💡 Hint

Look for the step where 'Queue empty, end traversal' is noted in execution_table.

If the tree had no right children, how would the queue state change at step 5?

A [2, 3]

B [3]

C [2]

D []

💡 Hint

At step 5, right child 3 is added to queue; if no right child, it won't be added.

Concept Snapshot

Left Side View of Binary Tree:
- Use level order traversal with a queue
- For each level, add the first node to the left view
- Traverse until queue is empty
- Return collected nodes as left side view
- Handles empty tree by returning empty list

Full Transcript

To find the left side view of a binary tree, we start at the root and use a queue to visit nodes level by level. At each level, we record the first node we see because it is the leftmost node visible from the left side. We add the left and right children of each node to the queue to process the next level. This continues until the queue is empty, meaning all nodes have been visited. If the tree is empty, we return an empty list immediately. The left view is the list of first nodes from 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