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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of the postorder traversal?

Given the binary tree below, what is the output of the postorder traversal (Left, Right, Root)?

Tree structure:
1
/ \
2 3
/ \
4 5

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function postorder(root) {
  if (!root) return [];
  return [...postorder(root.left), ...postorder(root.right), root.value];
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);

console.log(postorder(root));

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

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

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

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

💡 Hint

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

In postorder traversal, we visit left child, then right child, then the node itself. So for the tree, the order is 4 (left of 2), 5 (right of 2), 2 (root of left subtree), 3 (right child of root), then 1 (root).

❓ Predict Output

intermediate

What is the postorder traversal output for this tree?

Consider this binary tree:

10
/ \
20 30
/ \
40 50

What is the postorder traversal output (Left, Right, Root)?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

function postorder(root) {
  if (!root) return [];
  return [...postorder(root.left), ...postorder(root.right), root.value];
}

const root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.right.left = new Node(40);
root.right.right = new Node(50);

console.log(postorder(root));

A [20, 40, 50, 30, 10]

B [10, 20, 30, 40, 50]

C [40, 50, 30, 20, 10]

D [20, 30, 40, 50, 10]

💡 Hint

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

The traversal visits 20 (left child), then 40 and 50 (left and right of 30), then 30, and finally 10 (root).

🧠 Conceptual

advanced

Which statement about postorder traversal is true?

Choose the correct statement about postorder traversal of a binary tree.

A Postorder traversal visits the root node before its children.

B Postorder traversal visits left subtree, then right subtree, then the root node.

C Postorder traversal visits nodes in the order: Right, Left, Root.

D Postorder traversal visits nodes in the order: Root, Left, Right.

💡 Hint

Think about when the root node is visited in postorder.

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

🔧 Debug

advanced

Find the error in this postorder traversal code

The following code is intended to perform a postorder traversal of a binary tree and return an array of node values. What is the error?

DSA Javascript

function postorder(root) {
  if (!root) return [];
  return [root.value, ...postorder(root.left), ...postorder(root.right)];
}

A There is no error; this is correct postorder traversal.

B The function will cause a runtime error because of incorrect spread syntax.

C The function will return an empty array for any input.

D The root node is visited before its children, so this is preorder traversal, not postorder.

💡 Hint

Check the order in which root and children are visited.

The code visits the root first, then left and right children, which is preorder traversal, not postorder.

🚀 Application

expert

How many nodes are visited before the root in postorder traversal?

In a postorder traversal of a binary tree with 7 nodes, how many nodes are visited before the root node is visited?

A 6

B 7

C 0

D Depends on the tree structure

💡 Hint

Postorder visits all children before the root.

Postorder traversal visits all nodes in the left and right subtrees before visiting the root node, so all other nodes (6) are visited first.

Practice

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

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 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 A

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember: postorder ends with root node [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root node first instead of last
• Swapping left and right subtree order

2. Which of the following is the correct syntax to perform a postorder traversal function call in JavaScript assuming postorder(node) is defined?

easy

A. console.log(node.value); postorder(node.left); postorder(node.right);

B. console.log(node.value); postorder(node.right); postorder(node.left);

C. postorder(node.right); postorder(node.left); console.log(node.value);

D. postorder(node.left); postorder(node.right); console.log(node.value);

Solution

1. Step 1: Recall postorder traversal steps

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

2. Step 2: Check syntax order

   postorder(node.left); postorder(node.right); console.log(node.value); calls postorder on left, then right, then prints root value last, matching postorder.

3. Final Answer:

   postorder(node.left); postorder(node.right); console.log(node.value); -> Option D

4. Quick Check:

   Left, Right, Root call order = postorder(node.left); postorder(node.right); console.log(node.value); [OK]

Hint: Print root after left and right calls [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 a postorder traversal?

    1
   / \
  2   3
 / \
4   5

medium

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

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

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

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

Solution

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

   Left subtree rooted at 2: postorder is left(4), right(5), root(2) -> [4, 5, 2]

2. Step 2: Traverse right subtree and root

   Right subtree is node 3 -> [3]. Then root 1 last -> [1]. Combine: [4, 5, 2] + [3] + [1] = [4, 5, 2, 3, 1]

3. Final Answer:

   [4, 5, 2, 3, 1] -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Listing root before subtrees
• Mixing up left and right subtree order
• Confusing preorder or inorder output

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

function postorder(node) {
  if (!node) return;
  console.log(node.value);
  postorder(node.left);
  postorder(node.right);
}

medium

A. The left subtree is visited after the right subtree

B. The function does not check if node is null

C. The root node is printed before its children, not after

D. The function is missing a return statement

Solution

1. Step 1: Analyze print order in code

   The code prints node.value before recursive calls, so root prints first.

2. Step 2: Compare with postorder definition

   Postorder requires printing root after left and right subtrees, so printing first is wrong.

3. Final Answer:

   The root node is printed before its children, not after -> Option C

4. Quick Check:

   Postorder prints root last, code prints root first [OK]

Hint: Print root after recursive calls for postorder [OK]

Common Mistakes:

• Printing root before recursive calls
• Ignoring base case for null nodes
• Swapping left and right subtree calls

5. You want to collect all node values of a binary tree in postorder traversal into an array. Which code snippet correctly does this in JavaScript?

hard

A. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.left, arr);
postorder(node.right, arr);
arr.push(node.value);
return arr;
}

B. function postorder(node, arr = []) {
if (!node) return arr;
arr.push(node.value);
postorder(node.left, arr);
postorder(node.right, arr);
return arr;
}

C. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.right, arr);
postorder(node.left, arr);
arr.push(node.value);
return arr;
}

D. function postorder(node, arr = []) {
if (!node) return arr;
postorder(node.left, arr);
arr.push(node.value);
postorder(node.right, arr);
return arr;
}

Solution

1. Step 1: Check traversal order in each snippet

   function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } visits left subtree, then right subtree, then pushes root value last, matching postorder.

2. Step 2: Verify array accumulation and return

   function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } correctly passes and returns the array, accumulating values in postorder.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Left, Right, Root with arr.push last = function postorder(node, arr = []) { if (!node) return arr; postorder(node.left, arr); postorder(node.right, arr); arr.push(node.value); return arr; } [OK]

Hint: Push root value after left and right recursive calls [OK]

Common Mistakes:

• Pushing root value before recursive calls
• Swapping left and right subtree calls
• Not returning the array properly