Tree Traversal Postorder Left Right Root in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to call the postorder traversal function recursively on the left child.

DSA Go

func postorder(node *Node) {
    if node == nil {
        return
    }
    postorder(node.[1])
    // Continue traversal
}

Drag options to blanks, or click blank then click option'

ALeft

BRight

CParent

DRoot

Attempts:

3 left

💡 Hint

Common Mistakes

Calling the function on the right child first.

Calling the function on the node itself again.

Using incorrect field names.

In postorder traversal, we first visit the left child, so we call the function on node.Left.

2fill in blank

medium

Complete the code to call the postorder traversal function recursively on the right child.

DSA Go

func postorder(node *Node) {
    if node == nil {
        return
    }
    postorder(node.Left)
    postorder(node.[1])
    // Continue traversal
}

Drag options to blanks, or click blank then click option'

ARight

BRoot

CLeft

DParent

Attempts:

3 left

💡 Hint

Common Mistakes

Calling the function on the left child again.

Calling the function on the node itself instead of a child.

Using incorrect field names.

After visiting the left child, postorder traversal visits the right child, so we call the function on node.Right.

3fill in blank

hard

Fix the error in the code to print the node's value after visiting its children.

DSA Go

func postorder(node *Node) {
    if node == nil {
        return
    }
    postorder(node.Left)
    postorder(node.Right)
    fmt.Println(node.[1])
}

Drag options to blanks, or click blank then click option'

ALeft

BRight

CValue

DParent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the left or right child instead of the node's value.

Printing the parent pointer which may not exist.

Forgetting to print anything.

In postorder traversal, after visiting left and right children, we print the node's Value.

4fill in blank

hard

Fill both blanks to complete the postorder traversal function that returns a slice of node values.

DSA Go

func postorderValues(node *Node) []int {
    if node == nil {
        return []int{}
    }
    left := postorderValues(node.[1])
    right := postorderValues(node.[2])
    return append(append(left, right...), node.Value)
}

Drag options to blanks, or click blank then click option'

ALeft

BRight

CParent

DRoot

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children.

Using non-existent fields like Parent or Root.

Appending values in wrong order.

We recursively get values from the left child, then the right child, then add the current node's value.

5fill in blank

hard

Fill all three blanks to build a map of node values to their depths during postorder traversal.

DSA Go

func postorderDepthMap(node *Node, depth int, m map[int]int) {
    if node == nil {
        return
    }
    postorderDepthMap(node.[1], depth+1, m)
    postorderDepthMap(node.[2], depth+1, m)
    m[[3]] = depth
}

Drag options to blanks, or click blank then click option'

ALeft

BRight

Cnode.Value

Ddepth

Attempts:

3 left

💡 Hint

Common Mistakes

Using depth as map key instead of node value.

Swapping left and right children.

Not increasing depth during recursion.

We recurse on left and right children increasing depth, then store the node's value with its depth in the map.

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