Tree Traversal Postorder Left Right Root in DSA Java - 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 the order: Left subtree, Right subtree, then Root node.

intermediate

Why is Postorder traversal useful in deleting a tree?

Postorder visits children before the parent, so it deletes leaf nodes first, preventing dangling references.

beginner

Given a binary tree, what is the Postorder traversal output for this tree? 1 / \ 2 3 / \ 4 5

The Postorder traversal visits nodes as: 4, 5, 2, 3, 1

beginner

Show the recursive Java method signature for Postorder traversal of a binary tree.

public void postorderTraversal(TreeNode root) { if (root == null) return; postorderTraversal(root.left); postorderTraversal(root.right); System.out.print(root.val + " "); }

beginner

What is the main difference between Postorder and Preorder tree traversal?

Postorder visits Left, Right, then Root; Preorder visits Root, Left, then Right.

In Postorder traversal, which node is visited last?

A Leaf node

B Left child

C Right child

D Root node

Postorder visits Left, then Right, and finally the Root node.

What is the Postorder traversal of a tree with only one node (root)?

A Root node only

B Left child only

C No nodes visited

D Right child only

With one node, Postorder visits the root last, so it visits the root node.

Which traversal order is Postorder?

A Left, Right, Root

B Left, Root, Right

C Right, Left, Root

D Root, Left, Right

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

Which of these is a use case for Postorder traversal?

A Printing nodes in ascending order

B Deleting a tree safely

C Searching for a node quickly

D Building a tree from preorder data

Postorder traversal deletes children before parents, making it safe for tree deletion.

In Postorder traversal, what happens if the tree is empty?

A Traversal prints null

B Traversal prints root

C Traversal prints nothing

D Traversal throws error

If the tree is empty (null root), Postorder traversal does nothing and prints nothing.

Explain the steps of Postorder traversal on a binary tree.

Think about the order nodes are visited.

Write a simple recursive Java method for Postorder traversal and describe how it works.

Focus on the order of recursive calls and printing.

Practice

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

easy

A. Root node, Left subtree, Right subtree

B. Left subtree, Right subtree, Root node

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 the order Left, Right, Root exactly.

3. Final Answer:

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

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 Java method signatures correctly represents a recursive postorder traversal of a binary tree node named root?

easy

A. void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); }

B. void postorder(Node root) { visit(root); postorder(root.left); postorder(root.right); }

C. void postorder(Node root) { postorder(root.right); postorder(root.left); visit(root); }

D. void postorder(Node root) { visit(root.left); visit(root.right); visit(root); }

Solution

1. Step 1: Recall postorder traversal steps

   Postorder visits left subtree, then right subtree, then root node (visit).

2. Step 2: Check each method's order

   void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } calls postorder on left, then right, then visits root, matching postorder correctly.

3. Final Answer:

   void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } -> Option A

4. Quick Check:

   Left, Right, Root method order = void postorder(Node root) { postorder(root.left); postorder(root.right); visit(root); } [OK]

Hint: Postorder calls left, right, then visit root [OK]

Common Mistakes:

• Visiting root before children
• Swapping left and right recursive calls
• Using visit instead of recursive call for children

3. Given the binary tree below:

    1
   / \
  2   3
 / \
4   5

What is the output of the postorder traversal?

medium

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

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

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

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

Solution

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

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

2. Step 2: Traverse right subtree and root

   Right subtree is 3 (leaf), so visit 3, then visit root 1 -> [3, 1]

3. Step 3: Combine all parts

   Full postorder: left subtree [4,5,2], right subtree [3], root [1] -> [4, 5, 2, 3, 1]

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

• Visiting root before children
• Mixing up left and right subtree order
• Outputting preorder or inorder instead

4. Consider this Java method for postorder traversal:

void postorder(Node root) {
  if (root == null) return;
  postorder(root.right);
  postorder(root.left);
  visit(root);
}

What is the main problem with this code?

medium

A. It visits right subtree before left subtree, reversing order

B. It causes infinite recursion

C. It visits the root before children

D. It does not check for null root

Solution

1. Step 1: Analyze traversal order in code

   The code calls postorder on right subtree first, then left subtree, then visits root.

2. Step 2: Compare with correct postorder order

   Correct postorder is left, right, root. This code reverses left and right subtree order.

3. Final Answer:

   It visits right subtree before left subtree, reversing order -> Option A

4. Quick Check:

   Left before right needed, code does right before left [OK]

Hint: Left subtree must be visited before right in postorder [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Ignoring null check (though present here)
• Visiting root before children

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

hard

A. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); result.add(root.val); postorder(root.right, result); }

B. void postorder(Node root, List<Integer> result) { if(root==null) return; result.add(root.val); postorder(root.left, result); postorder(root.right, result); }

C. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.right, result); postorder(root.left, result); result.add(root.val); }

D. void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); }

Solution

1. Step 1: Understand postorder traversal with list collection

   Postorder visits left subtree, then right subtree, then adds root value to list.

2. Step 2: Check each code snippet's order

   void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } calls postorder on left, then right, then adds root value last, matching postorder correctly.

3. Final Answer:

   void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } -> Option D

4. Quick Check:

   Left, Right, Root with result.add last = void postorder(Node root, List<Integer> result) { if(root==null) return; postorder(root.left, result); postorder(root.right, result); result.add(root.val); } [OK]

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

Common Mistakes:

• Adding root value before children
• Swapping left and right subtree calls
• Forgetting null check before recursion