Tree Traversal Preorder Root Left Right in DSA Go - Time & Space Complexity

Time Complexity: Tree Traversal Preorder Root Left Right

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

Specifically, we ask: How does preorder traversal scale with the number of nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func preorder(root *Node) {
    if root == nil {
        return
    }
    fmt.Println(root.value) // Visit root
    preorder(root.left)     // Traverse left subtree
    preorder(root.right)    // Traverse right subtree
}
    

This code visits each node starting from the root, then left child, then right child recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once (printing value).
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the number of visits grows at the same rate.

Input Size (n)	Approx. Operations
10	10 visits
100	100 visits
1000	1000 visits

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete preorder traversal grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Preorder traversal takes more than linear time because it visits nodes multiple times."

[OK] Correct: Each node is visited exactly once, so the total work is proportional to the number of nodes, not more.

Interview Connect

Understanding preorder traversal time helps you explain how tree algorithms scale and shows you can analyze recursive code clearly.

Self-Check

"What if we changed preorder to inorder traversal? How would the time complexity change?"

Practice

1. What is the order of nodes visited in a preorder tree traversal?

easy

A. Root, Left, Right

B. Left, Root, Right

C. Left, Right, Root

D. Right, Left, Root

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then the left subtree, and finally the right subtree.

2. Step 2: Match the order with options

   Root, Left, Right matches the order Root, Left, Right exactly.

3. Final Answer:

   Root, Left, Right -> Option A

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Remember preorder starts with root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Swapping left and right subtree order
• Starting traversal from left child instead of root

2. Which of the following Go function signatures correctly represents a preorder traversal function for a binary tree node Node?

easy

A. func preorder(root *Node) { if root == nil { return } preorder(root.Left) preorder(root.Right) fmt.Println(root.Value) }

B. func preorder(root *Node) { if root == nil { return } fmt.Println(root.Value) preorder(root.Left) preorder(root.Right) }

C. func preorder(root *Node) { if root == nil { return } preorder(root.Right) fmt.Println(root.Value) preorder(root.Left) }

D. func preorder(root *Node) { if root == nil { return } preorder(root.Left) fmt.Println(root.Value) preorder(root.Right) }

Solution

1. Step 1: Recall preorder traversal order

   Preorder visits root first, then left subtree, then right subtree.

2. Step 2: Check function order

   func preorder(root *Node) { if root == nil { return } fmt.Println(root.Value) preorder(root.Left) preorder(root.Right) } prints root value first, then recursively visits left and right children, matching preorder.

3. Final Answer:

   func preorder(root *Node) { if root == nil { return } fmt.Println(root.Value) preorder(root.Left) preorder(root.Right) } -> Option B

4. Quick Check:

   Print root before children calls [OK]

Hint: Print root value before recursive calls [OK]

Common Mistakes:

• Printing after recursive calls (postorder)
• Swapping left and right recursive calls
• Missing nil check for root

3. Given the binary tree below, what is the preorder traversal output?

    10
   /  \
  5    15
 /    /  \
3    12  20

func preorder(root *Node) { if root == nil { return } fmt.Print(root.Value, " ") preorder(root.Left) preorder(root.Right) }

medium

A. [5, 3, 10, 15, 12, 20]

B. [3, 5, 10, 12, 15, 20]

C. [10, 15, 20, 12, 5, 3]

D. [10, 5, 3, 15, 12, 20]

Solution

1. Step 1: Start at root node 10

   Print 10, then traverse left subtree.

2. Step 2: Traverse left subtree (5, then 3)

   Print 5, then left child 3, which has no children.

3. Step 3: Traverse right subtree (15, 12, 20)

   Print 15, then left child 12, then right child 20.

4. Final Answer:

   [10, 5, 3, 15, 12, 20] -> Option D

5. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Follow root, then left subtree, then right subtree order [OK]

Common Mistakes:

• Mixing preorder with inorder or postorder output
• Visiting right subtree before left
• Skipping leaf nodes

4. The following Go code attempts a preorder traversal but produces incorrect output. What is the error?

func preorder(root *Node) {
  if root == nil {
    return
  }
  preorder(root.Left)
  fmt.Print(root.Value, " ")
  preorder(root.Right)
}

medium

A. The function uses fmt.Print instead of fmt.Println

B. The function does not check for nil root

C. The root value is printed after left subtree traversal, not before

D. The right subtree is traversed before the left subtree

Solution

1. Step 1: Analyze print position in code

   The root value is printed after preorder(root.Left), so root is visited after left subtree.

2. Step 2: Compare with preorder definition

   Preorder requires visiting root first, then left, then right. This code visits left first, then root, then right (inorder).

3. Final Answer:

   The root value is printed after left subtree traversal, not before -> Option C

4. Quick Check:

   Print root before children for preorder [OK]

Hint: Print root value before recursive calls to fix order [OK]

Common Mistakes:

• Confusing preorder with inorder traversal
• Ignoring nil check (though present here)
• Thinking print function choice affects traversal order

5. You want to collect all node values of a binary tree in preorder traversal into a slice in Go. Which code snippet correctly does this?

hard

A. func preorder(root *Node, result *[]int) { if root == nil { return } *result = append(*result, root.Value) preorder(root.Left, result) preorder(root.Right, result) }

B. func preorder(root *Node, result []int) []int { if root == nil { return result } result = append(result, root.Value) preorder(root.Left, result) preorder(root.Right, result) return result }

C. func preorder(root *Node) []int { if root == nil { return nil } left := preorder(root.Left) right := preorder(root.Right) return append(left, append(right, root.Value)...) }

D. func preorder(root *Node, result *[]int) { if root == nil { return } preorder(root.Left, result) preorder(root.Right, result) *result = append(*result, root.Value) }

Solution

1. Step 1: Understand preorder collection logic

   Preorder means add root value first, then traverse left and right subtrees.

2. Step 2: Check each option's order and parameter usage

   func preorder(root *Node, result *[]int) { if root == nil { return } *result = append(*result, root.Value) preorder(root.Left, result) preorder(root.Right, result) } appends root value first, then recursively traverses left and right, using pointer to slice to collect values correctly.

3. Final Answer:

   func preorder(root *Node, result *[]int) { if root == nil { return } *result = append(*result, root.Value) preorder(root.Left, result) preorder(root.Right, result) } -> Option A

4. Quick Check:

   Append root first, use pointer to slice [OK]

Hint: Append root value before recursive calls, pass slice pointer [OK]

Common Mistakes:

• Appending after recursive calls (wrong order)
• Passing slice by value and expecting changes
• Appending right subtree before left subtree