Tree Traversal Postorder Left Right Root in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Postorder Traversal on a Simple Binary Tree

What is the output of the postorder traversal (Left, Right, Root) for the given binary tree?

DSA Go

package main

import "fmt"

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

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

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

    var result []int
    postorder(root, &result)
    fmt.Println(result)
}

A [2, 4, 5, 3, 1]

B [1, 2, 3, 4, 5]

C [4, 2, 5, 3, 1]

D [4, 5, 2, 3, 1]

💡 Hint

Remember postorder means visit left subtree, then right subtree, then the root node.

In postorder traversal, we first visit the left child nodes, then the right child nodes, and finally the root node. For this tree, the order is 4 (leftmost leaf), 5 (right child of 2), 2 (left child of root), 3 (right child of root), and finally 1 (root).

🧠 Conceptual

intermediate

Understanding Postorder Traversal Steps

Which of the following best describes the order of visiting nodes in postorder traversal?

A Left, Right, Root

B Root, Left, Right

C Right, Left, Root

D Left, Root, Right

💡 Hint

Postorder traversal always visits the root last.

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

🔧 Debug

advanced

Identify the Bug in Postorder Traversal Implementation

What error will this code produce when performing postorder traversal?

DSA Go

package main

import "fmt"

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

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

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

    var result []int
    postorder(root, &result)
    fmt.Println(result)
}

A Output: [3, 2, 1] but order is incorrect for postorder

B Output: [3, 2, 1]

C Output: [2, 3, 1]

D Runtime error: nil pointer dereference

💡 Hint

Check the order of recursive calls for left and right children.

The code visits the right subtree first, then the left subtree, then the root. This reverses the left and right order, so the output is not a correct postorder traversal.

🚀 Application

advanced

Postorder Traversal Output for Unbalanced Tree

Given the following unbalanced binary tree, what is the postorder traversal output?

DSA Go

package main

import "fmt"

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

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

func main() {
    root := &Node{Val: 10}
    root.Right = &Node{Val: 20}
    root.Right.Left = &Node{Val: 15}
    root.Right.Right = &Node{Val: 25}

    var result []int
    postorder(root, &result)
    fmt.Println(result)
}

A [15, 20, 25, 10]

B [10, 15, 25, 20]

C [15, 25, 20, 10]

D [20, 15, 25, 10]

💡 Hint

Remember to visit left subtree, then right subtree, then root.

The left child of 20 is 15, right child is 25. Postorder visits 15, then 25, then 20, then finally 10.

🧠 Conceptual

expert

Number of Nodes Visited in Postorder Traversal

If a binary tree has 7 nodes, how many nodes will be visited during a complete postorder traversal?

A 6

B 7

C 8

D Depends on the tree structure

💡 Hint

Postorder traversal visits every node exactly once.

Postorder traversal visits every node in the tree exactly once, so the number of nodes visited equals the total number of nodes.

Practice

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

easy

A. Root node, Right subtree, Left subtree

B. Root node, Left subtree, Right subtree

C. Left subtree, Right subtree, Root node

D. Right subtree, Left subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

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

2. Step 2: Match the order with options

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

3. Final Answer:

   Left subtree, Right subtree, Root node -> Option C

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember postorder ends with root after children [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root before children
• Swapping left and right subtree order

2. Which of the following Go function signatures correctly represents a recursive postorder traversal of a binary tree node root?

easy

A. func postorder(root *Node) { postorder(root.Left); postorder(root.Right); fmt.Println(root.Value) }

B. func postorder(root *Node) { fmt.Println(root.Value); postorder(root.Left); postorder(root.Right) }

C. func postorder(root *Node) { postorder(root.Right); postorder(root.Left); fmt.Println(root.Value) }

D. func postorder(root *Node) { fmt.Println(root.Value); postorder(root.Right); postorder(root.Left) }

Solution

1. Step 1: Recall postorder traversal steps

   Postorder visits left subtree, then right subtree, then root node.

2. Step 2: Check function calls and print order

   func postorder(root *Node) { postorder(root.Left); postorder(root.Right); fmt.Println(root.Value) } calls postorder on left, then right, then prints root value, matching postorder.

3. Final Answer:

   func postorder(root *Node) { postorder(root.Left); postorder(root.Right); fmt.Println(root.Value) } -> Option A

4. Quick Check:

   Calls: Left, Right, Print root [OK]

Hint: Print root after recursive calls to left and right [OK]

Common Mistakes:

• Printing root before recursive calls
• Swapping left and right recursive calls
• Missing recursive calls for subtrees

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

      1
     / \
    2   3
   / \
  4   5

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

medium

A. 4 2 5 3 1

B. 1 2 4 5 3

C. 2 4 5 3 1

D. 4 5 2 3 1

Solution

1. Step 1: Traverse left subtree of root (1)

   Left subtree is rooted at 2. Postorder of subtree 2 is: left(4), right(5), root(2) -> 4 5 2.

2. Step 2: Traverse right subtree and root

   Right subtree is 3 (leaf), so output 3. Then print root 1 last.

3. Final Answer:

   4 5 2 3 1 -> Option D

4. Quick Check:

   Postorder = Left, Right, Root = 4 5 2 3 1 [OK]

Hint: Visit left subtree fully, then right, then root [OK]

Common Mistakes:

• Printing root before children
• Mixing left and right subtree order
• Skipping leaf nodes

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

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

medium

A. The order of recursive calls is swapped; left should be before right

B. The base case is missing for nil nodes

C. The root value should be printed before recursive calls

D. The function should return a slice instead of printing

Solution

1. Step 1: Analyze recursive call order

   The code calls postorder on right subtree first, then left subtree, which reverses the expected order.

2. Step 2: Recall correct postorder sequence

   Postorder requires left subtree first, then right subtree, then root. Swapping left and right causes wrong output.

3. Final Answer:

   The order of recursive calls is swapped; left should be before right -> Option A

4. Quick Check:

   Left before Right in postorder [OK]

Hint: Left subtree call must come before right subtree call [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Printing root before children
• Ignoring base case for nil

5. You want to collect the values of a binary tree in postorder traversal into a slice instead of printing. Which Go function correctly implements this?

hard

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

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

C. func postorder(root *Node, res *[]int) {
if root == nil {
return
}
*res = append(*res, root.Value)
postorder(root.Left, res)
postorder(root.Right, res)
}

D. func postorder(root *Node) []int {
if root == nil {
return nil
}
return []int{root.Value}
}

Solution

1. Step 1: Understand collecting values in slice recursively

   We want a function returning a slice with postorder values: left subtree slice + right subtree slice + root value.

2. Step 2: Check each option for correct order and slice handling

   func postorder(root *Node) []int { if root == nil { return []int{} } left := postorder(root.Left) right := postorder(root.Right) return append(left, append(right, root.Value)...) } correctly returns left slice, then appends right slice, then root value, matching postorder traversal.

3. Final Answer:

   func postorder(root *Node) []int { if root == nil { return []int{} } left := postorder(root.Left) right := postorder(root.Right) return append(left, append(right, root.Value)...) } -> Option B

4. Quick Check:

   Return slices: left + right + root [OK]

Hint: Return left + right + root slices combined [OK]

Common Mistakes:

• Appending root before children slices
• Not combining slices correctly
• Returning nil instead of empty slice for nil node