Tree Traversal Postorder Left Right Root in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to visit the left child first in postorder traversal.

DSA Java

if (node.left != null) {
    [1];
}

Drag options to blanks, or click blank then click option'

Apostorder(node.left)

Bpreorder(node.left)

Cinorder(node.left)

DlevelOrder(node.left)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder or inorder instead of postorder.

Not calling the traversal on the left child.

In postorder traversal, we visit the left child first by calling postorder(node.left).

2fill in blank

medium

Complete the code to visit the right child second in postorder traversal.

DSA Java

if (node.right != null) {
    [1];
}

Drag options to blanks, or click blank then click option'

Apreorder(node.right)

Binorder(node.right)

Cpostorder(node.right)

DlevelOrder(node.right)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder or inorder instead of postorder.

Not calling the traversal on the right child.

In postorder traversal, after visiting the left child, we visit the right child by calling postorder(node.right).

3fill in blank

hard

Fix the error in printing the root node's value in postorder traversal.

DSA Java

System.out.println([1]);

Drag options to blanks, or click blank then click option'

Anode.left

Bnode

Cnode.right

Dnode.value

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the node object instead of its value.

Printing left or right child instead of root.

To print the root node's value, use node.value. Printing node or its children will not print the value.

4fill in blank

hard

Fill both blanks to complete the recursive postorder traversal method.

DSA Java

public void postorder(Node node) {
    if (node == null) return;
    [1];
    [2];
    System.out.println(node.value);
}

Drag options to blanks, or click blank then click option'

Apostorder(node.left)

Bpreorder(node.left)

Cpostorder(node.right)

Dinorder(node.right)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling preorder or inorder instead of postorder.

Printing before visiting children.

Postorder traversal visits left child first, then right child, then root. So call postorder(node.left) and postorder(node.right).

5fill in blank

hard

Fill all three blanks to create a postorder traversal that collects values in a list.

DSA Java

public void postorderCollect(Node node, List<Integer> result) {
    if (node == null) return;
    [1];
    [2];
    result.[3](node.value);
}

Drag options to blanks, or click blank then click option'

ApostorderCollect(node.left, result)

BpostorderCollect(node.right, result)

Cadd

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using remove instead of add on the list.

Not passing the list in recursive calls.

To collect values in postorder, recursively call postorderCollect on left and right children, then add the root value to the list using add.

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