Tree Traversal Postorder Left Right Root in DSA Go

Mental Model

Visit the left child, then the right child, and finally the node itself. This order ensures children are processed before their parent.

Analogy: Imagine cleaning a tree house: first clean the left room, then the right room, and only after both rooms are clean, clean the main hall.

    1
   / \
  2   3
 / \
4   5

↑ root at 1

Dry Run Walkthrough

Input: Tree: 1 with left child 2 and right child 3; 2 has children 4 (left) and 5 (right)

Goal: Print nodes in postorder: left subtree, right subtree, then root

Step 1: Visit left subtree of 1: move to node 2

1 -> [curr->2] -> 3
2 -> 4 -> 5

Why: We must process left child before root

Step 2: Visit left subtree of 2: move to node 4

2 -> [curr->4] -> 5

Why: Continue left before right or root

Step 3: Node 4 has no children, print 4

Printed: 4

Why: Leaf node reached, print now

Step 4: Visit right subtree of 2: move to node 5

2 -> 4 -> [curr->5]

Why: After left subtree, process right subtree

Step 5: Node 5 has no children, print 5

Printed: 4 5

Why: Leaf node reached, print now

Step 6: Print node 2 after its children

Printed: 4 5 2

Why: Children processed, now print parent

Step 7: Visit right subtree of 1: move to node 3

1 -> 2 -> [curr->3]

Why: After left subtree, process right subtree

Step 8: Node 3 has no children, print 3

Printed: 4 5 2 3

Why: Leaf node reached, print now

Step 9: Print root node 1 after its children

Printed: 4 5 2 3 1

Why: All children processed, print root last

Result:

4 5 2 3 1

Annotated Code

DSA Go

package main

import "fmt"

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

func postorderTraversal(root *TreeNode) {
    if root == nil {
        return
    }
    postorderTraversal(root.Left) // traverse left subtree first
    postorderTraversal(root.Right) // then traverse right subtree
    fmt.Printf("%d ", root.Val) // print root after children
}

func main() {
    // Construct the tree:
    //     1
    //    / \
    //   2   3
    //  / \
    // 4   5
    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}

    postorderTraversal(root)
    fmt.Println()
}

postorderTraversal(root.Left) // traverse left subtree first

advance to left child to process left subtree

postorderTraversal(root.Right) // then traverse right subtree

after left subtree, process right subtree

fmt.Printf("%d ", root.Val) // print root after children

print current node after children processed

Output

4 5 2 3 1 

Complexity Analysis

Time: O(n) because each node is visited exactly once

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Compared to iterative traversal, recursive is simpler but uses call stack; iterative may use explicit stack with similar time

Edge Cases

Empty tree (root is nil)

Function returns immediately without printing

DSA Go

if root == nil {
        return
    }

Single node tree

Prints the single node value

DSA Go

if root == nil {
        return
    }

When to Use This Pattern

When you need to process children before their parent in a tree, use postorder traversal to ensure left and right subtrees are handled first.

Common Mistakes

Mistake: Printing the root before children (preorder) instead of after
Fix: Move the print statement after recursive calls to left and right children

Summary

Visits left subtree, then right subtree, then the node itself.

Use when you need to process children before their parent, like deleting a tree.

The key is to print the node only after both children are processed.

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