Tree Traversal Postorder Left Right Root in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the order of visiting nodes in Postorder Tree Traversal?

In Postorder traversal, nodes are visited in this order: Left subtree, Right subtree, then Root node.

intermediate

Why is Postorder traversal useful in expression trees?

Postorder traversal visits children before the parent, which matches the order needed to evaluate expressions (operands before operators).

beginner

Show the Postorder traversal output for this tree:
Root: 1
Left child: 2
Right child: 3

The Postorder traversal visits Left (2), Right (3), then Root (1). So output is: 2 -> 3 -> 1

beginner

What is the base case in a recursive Postorder traversal function?

The base case is when the current node is null (no node), then return without doing anything.

beginner

How does Postorder traversal differ from Preorder traversal?

Postorder visits Left, Right, Root; Preorder visits Root, Left, Right. The root is visited last in Postorder but first in Preorder.

In Postorder traversal, which node is visited last?

A Root node

B Left child

C Right child

D None of the above

Postorder visits Left, then Right, then Root last.

What is the first step in Postorder traversal of a node?

A Visit the root node

B Visit the right subtree

C Return null

D Visit the left subtree

Postorder starts by visiting the left subtree first.

Which traversal order is Postorder?

A Left, Right, Root

B Root, Left, Right

C Left, Root, Right

D Right, Left, Root

Postorder visits Left subtree, then Right subtree, then Root.

Postorder traversal is especially useful for which of these?

A Finding the height of a tree

B Printing nodes in ascending order

C Evaluating expression trees

D Searching for a value

Postorder visits children before parent, matching expression evaluation order.

What happens if the node is null in Postorder traversal?

A Traverse left subtree

B Return immediately

C Visit the node

D Traverse right subtree

If node is null, the function returns immediately (base case).

Explain how Postorder traversal visits nodes in a binary tree.

Think about visiting children before the parent.

Describe a real-life example where Postorder traversal is useful.

Consider tasks that require finishing children before parent.

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