Tree Traversal Postorder Left Right Root in DSA Rust - Execution Trace

Concept Flow - Tree Traversal Postorder Left Right Root

Start at root node

↓

Traverse left subtree

↓

Traverse right subtree

↓

Visit current node (root)

↓

Back to parent node or end

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

Execution Sample

DSA Rust

fn postorder(node: &Option<Box<TreeNode>>) {
    if let Some(n) = node {
        postorder(&n.left);
        postorder(&n.right);
        println!("{}", n.val);
    }
}

This code recursively visits left child, right child, then prints the current node value.

Execution Table

Step	Operation	Node Visited	Left/Right	Tree State
1	Start at root	1	root	1 ├─2 └─3
2	Traverse left subtree	2	left of 1	1 ├─2 └─3
3	Traverse left subtree	4	left of 2	1 ├─2 │ ├─4 │ └─5 └─3
4	Visit node	4	leaf	4
5	Traverse right subtree	5	right of 2	1 ├─2 │ ├─4 │ └─5 └─3
6	Visit node	5	leaf	5
7	Visit node	2	left subtree root	2 ├─4 └─5
8	Traverse right subtree	3	right of 1	1 ├─2 └─3 ├─6 └─7
9	Traverse left subtree	6	left of 3	3 ├─6 └─7
10	Visit node	6	leaf	6
11	Traverse right subtree	7	right of 3	3 ├─6 └─7
12	Visit node	7	leaf	7
13	Visit node	3	right subtree root	3 ├─6 └─7
14	Visit node	1	root	1 ├─2 └─3
15	End traversal	-	-	-

💡 All nodes visited in postorder: left, right, root.

Variable Tracker

Variable	Start	After Step 2	After Step 7	After Step 8	After Step 13	Final
Current Node	1	2	1	3	1	-
Visited Nodes	[]	[]	[4,5,2]	[4,5,2]	[4,5,2,6,7,3]	[4,5,2,6,7,3,1]
Call Stack Depth	1	2	1	2	1	0

Key Moments - 3 Insights

Why do we visit the root node last in postorder traversal?

Because postorder means visiting left subtree, then right subtree, and only then the root node, as shown in steps 14 and 15 in the execution_table.

What happens when a node has no left or right child?

The traversal visits the node immediately since left and right subtrees are empty, like nodes 4, 5, 6, and 7 in steps 4, 6, 10, and 12.

How does the call stack change during recursion?

The call stack depth increases when going deeper into left or right children and decreases after visiting nodes, as tracked in variable_tracker under Call Stack Depth.

Visual Quiz - 3 Questions

Look at the execution_table, what node is visited at step 7?

A Node 4

B Node 3

C Node 2

D Node 5

💡 Hint

Check the 'Node Visited' column at step 7 in execution_table.

At which step does the traversal visit the root node?

A Step 14

B Step 13

C Step 15

D Step 12

💡 Hint

Look for the root node (1) visit in the 'Node Visited' column.

If node 5 had a left child, how would the execution_table change?

A Node 5 would be visited earlier

B More steps would appear before visiting node 5

C The root node would be visited first

D No change in steps

💡 Hint

Adding a child means more traversal steps before visiting that node, check how left subtree traversal works.

Concept Snapshot

Postorder traversal visits nodes in Left → Right → Root order.
It uses recursion to go deep left, then right, then processes the node.
Commonly used for deleting trees or postfix expressions.
In code, call postorder(left), postorder(right), then visit node.
Traversal ends when all nodes are visited.

Full Transcript

Postorder tree traversal means visiting the left subtree first, then the right subtree, and finally the root node. The process uses recursion to go as deep as possible on the left side, then the right side, and only then processes the current node. This is shown in the execution table where nodes 4 and 5 are visited before node 2, and nodes 6 and 7 before node 3, and finally node 1. The call stack grows as the recursion goes deeper and shrinks as nodes are visited. This traversal is useful for operations like deleting a tree or evaluating postfix expressions.

Practice

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

easy

A. Left subtree, Right subtree, Root node

B. Root node, Left subtree, Right subtree

C. Root node, Right subtree, Left subtree

D. Right subtree, Left subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

   Postorder traversal visits nodes in the order: left subtree first, then right subtree, and finally the root node.

2. Step 2: Match the order with options

   Left subtree, Right subtree, Root node matches this order exactly: Left subtree, Right subtree, Root node.

3. Final Answer:

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

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 Rust function signatures correctly represents a recursive postorder traversal function for a binary tree node defined as struct Node { left: Option, right: Option, value: i32 }?

easy

A. fn postorder(node: &Option) { /* visit left, right, root */ }

B. fn postorder(node: Node) { /* visit root, left, right */ }

C. fn postorder(node: &Node) -> i32 { /* visit root only */ }

D. fn postorder(node: &mut Node) -> bool { /* visit right, left, root */ }

Solution

1. Step 1: Identify correct parameter type for recursion

   Postorder traversal usually takes a reference to an optional boxed node to handle empty children safely.

2. Step 2: Match function signature with traversal order and parameter

   fn postorder(node: &Option) { /* visit left, right, root */ } uses &Option which is correct for recursive calls and mentions visiting left, right, root.

3. Final Answer:

   fn postorder(node: &Option) { /* visit left, right, root */ } -> Option A

4. Quick Check:

   Use &Option for safe recursion [OK]

Hint: Use &Option for recursive tree traversal [OK]

Common Mistakes:

• Passing ownership instead of reference
• Returning incorrect types
• Wrong traversal order in comments

3. Given the following Rust code for postorder traversal, what will be printed?

struct Node {
    value: i32,
    left: Option,
    right: Option,
}

fn postorder(node: &Option) {
    if let Some(n) = node {
        postorder(&n.left);
        postorder(&n.right);
        print!("{} ", n.value);
    }
}

fn main() {
    let tree = Some(Box::new(Node {
        value: 1,
        left: Some(Box::new(Node {
            value: 2,
            left: None,
            right: None,
        })),
        right: Some(Box::new(Node {
            value: 3,
            left: None,
            right: None,
        })),
    }));
    postorder(&tree);
}

medium

A. 1 3 2

B. 1 2 3

C. 3 2 1

D. 2 3 1

Solution

1. Step 1: Trace postorder traversal on given tree

   The tree root is 1, left child 2, right child 3. Postorder visits left (2), right (3), then root (1).

2. Step 2: Confirm printed order

   Values printed in order: 2, then 3, then 1, matching 2 3 1 .

3. Final Answer:

   2 3 1 -> Option D

4. Quick Check:

   Postorder prints left, right, root = 2 3 1 [OK]

Hint: Print order matches left-right-root traversal [OK]

Common Mistakes:

• Confusing preorder or inorder output
• Missing spaces in print output
• Ignoring None children

4. The following Rust code attempts a postorder traversal but has a bug. What is the error?

fn postorder(node: &Option) {
    if let Some(n) = node {
        print!("{} ", n.value);
        postorder(&n.left);
        postorder(&n.right);
    }
}

medium

A. It causes a stack overflow due to infinite recursion

B. It prints root before children, not postorder

C. It does not handle None nodes correctly

D. It borrows node mutably but parameter is immutable

Solution

1. Step 1: Analyze print statement position

   The print happens before recursive calls to left and right children, so root is printed first.

2. Step 2: Compare with postorder definition

   Postorder requires visiting left and right children before root, so this order is incorrect.

3. Final Answer:

   It prints root before children, not postorder -> Option B

4. Quick Check:

   Print after children for postorder [OK]

Hint: Print root after children for correct postorder [OK]

Common Mistakes:

• Printing root before recursive calls
• Confusing preorder with postorder
• Ignoring None checks

5. You want to collect all node values of a binary tree in postorder traversal into a vector in Rust. Which of the following code snippets correctly implements this?

hard

A. fn postorder_collect(node: &Option) -> Vec<i32> {
let mut vals = Vec::new();
if let Some(n) = node {
vals.push(n.value);
vals.extend(postorder_collect(&n.left));
vals.extend(postorder_collect(&n.right));
}
vals
}

B. fn postorder_collect(node: &Option, vals: &mut Vec<i32>) {
if let Some(n) = node {
vals.push(n.value);
postorder_collect(&n.left, vals);
postorder_collect(&n.right, vals);
}
}

C. fn postorder_collect(node: &Option) -> Vec<i32> {
let mut vals = Vec::new();
if let Some(n) = node {
vals.extend(postorder_collect(&n.left));
vals.extend(postorder_collect(&n.right));
vals.push(n.value);
}
vals
}

D. fn postorder_collect(node: &Option, vals: &mut Vec<i32>) {
if let Some(n) = node {
postorder_collect(&n.right, vals);
postorder_collect(&n.left, vals);
vals.push(n.value);
}
}

Solution

1. Step 1: Understand postorder collection logic

   Postorder means collect left subtree values, then right subtree values, then root value.

2. Step 2: Check each code snippet for correct order and return type

   fn postorder_collect(node: &Option) -> Vec<i32> { let mut vals = Vec::new(); if let Some(n) = node { vals.extend(postorder_collect(&n.left)); vals.extend(postorder_collect(&n.right)); vals.push(n.value); } vals } correctly collects left and right subtree vectors first, then pushes root value, returning the combined vector.

3. Final Answer:

   fn postorder_collect(node: &Option) -> Vec<i32> { let mut vals = Vec::new(); if let Some(n) = node { vals.extend(postorder_collect(&n.left)); vals.extend(postorder_collect(&n.right)); vals.push(n.value); } vals } -> Option C

4. Quick Check:

   Collect left, right, then root in vector [OK]

Hint: Collect children first, then push root value [OK]

Common Mistakes:

• Pushing root before children
• Not returning combined vector
• Using mutable vector incorrectly