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DSA C#programming~10 mins

Create a Binary Tree Manually in DSA C# - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a new binary tree node with value 10.

DSA C#
TreeNode root = new TreeNode([1]);
Drag options to blanks, or click blank then click option'
A10
B"10"
Cnull
Droot
Attempts:
3 left
💡 Hint
Common Mistakes
Using quotes around 10 makes it a string, which is incorrect.
Using null will create an empty node, not a node with value.
2fill in blank
medium

Complete the code to assign a left child node with value 5 to the root.

DSA C#
root.[1] = new TreeNode(5);
Drag options to blanks, or click blank then click option'
Aright
Bnode
Cchild
Dleft
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'right' assigns to the wrong child.
Using 'child' or 'node' are not valid properties.
3fill in blank
hard

Fix the error in the code to assign a right child node with value 15 to the root.

DSA C#
root.[1] = new TreeNode(15);
Drag options to blanks, or click blank then click option'
Aright
BleftChild
CRight
DchildRight
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'Right' with uppercase R causes a compile error.
Using 'leftChild' or 'childRight' are invalid property names.
4fill in blank
hard

Fill both blanks to create a root node with value 20 and assign a left child with value 10.

DSA C#
TreeNode root = new TreeNode([1]);
root.[2] = new TreeNode(10);
Drag options to blanks, or click blank then click option'
A20
Bleft
Cright
D15
Attempts:
3 left
💡 Hint
Common Mistakes
Assigning the child to 'right' instead of 'left'.
Using wrong root value.
5fill in blank
hard

Fill all three blanks to create a root node with value 30, assign left child 20 and right child 40.

DSA C#
TreeNode root = new TreeNode([1]);
root.[2] = new TreeNode(20);
root.[3] = new TreeNode(40);
Drag options to blanks, or click blank then click option'
A30
Bleft
Cright
D50
Attempts:
3 left
💡 Hint
Common Mistakes
Mixing up left and right properties.
Using wrong values for nodes.

Practice

(1/5)
1. What is the correct way to create a root node manually in a binary tree using C#?
easy
A. Node root = new Node(); root = 10;
B. Node root = Node(10);
C. Node root = new Node(); root.value = 10;
D. Node root = new Node(10);

Solution

  1. Step 1: Understand Node creation syntax

    In C#, to create an object, use the 'new' keyword followed by the constructor call with parameters.
  2. Step 2: Apply correct syntax for root node

    Using 'new Node(10)' correctly creates a node with value 10 assigned.
  3. Final Answer:

    Node root = new Node(10); -> Option D
  4. Quick Check:

    Use 'new' keyword with constructor to create node [OK]
Hint: Use 'new' keyword with constructor to create nodes [OK]
Common Mistakes:
  • Omitting 'new' keyword when creating node
  • Assigning value directly to node variable
  • Calling constructor without 'new'
2. Which of the following is the correct syntax to assign a left child node with value 5 to a root node named root?
easy
A. root.left = new Node(5);
B. root.left = Node(5);
C. root.left.value = 5;
D. root.left = 5;

Solution

  1. Step 1: Understand child node assignment

    Left child must be assigned a new Node object, not just a value.
  2. Step 2: Use correct syntax to assign left child

    Assigning 'root.left = new Node(5);' creates a new node with value 5 as left child.
  3. Final Answer:

    root.left = new Node(5); -> Option A
  4. Quick Check:

    Assign new Node object to left child [OK]
Hint: Assign new Node object, not just value, to child [OK]
Common Mistakes:
  • Assigning value directly instead of Node object
  • Omitting 'new' keyword
  • Trying to assign value to left property directly
3. Given the following code, what is the output of an inorder traversal (left, root, right)?
Node root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(3);
medium
A. [10, 5, 3, 15]
B. [3, 5, 10, 15]
C. [5, 3, 10, 15]
D. [3, 10, 5, 15]

Solution

  1. Step 1: Understand inorder traversal order

    Inorder visits left subtree, then root, then right subtree.
  2. Step 2: Trace traversal on given tree

    Left subtree of root (10) is node 5, which has left child 3. So traversal order is 3, 5, 10, 15.
  3. Final Answer:

    [3, 5, 10, 15] -> Option B
  4. Quick Check:

    Inorder = left, root, right [OK]
Hint: Inorder = left subtree, root, then right subtree [OK]
Common Mistakes:
  • Confusing preorder or postorder with inorder
  • Missing leftmost node in traversal
  • Listing nodes in insertion order
4. Identify the error in the following code that tries to create a binary tree:
Node root = new Node(20);
root.left = 10;
root.right = new Node(30);
medium
A. Right child node value is incorrect
B. Missing 'new' keyword when creating root node
C. Assigning integer directly to 'left' instead of a Node object
D. Root node value should be a string, not integer

Solution

  1. Step 1: Check left child assignment

    Left child must be a Node object, but code assigns integer 10 directly.
  2. Step 2: Verify other parts

    Root node and right child are correctly created with 'new Node()'.
  3. Final Answer:

    Assigning integer directly to 'left' instead of a Node object -> Option C
  4. Quick Check:

    Child nodes must be Node objects [OK]
Hint: Child nodes must be Node objects, not raw values [OK]
Common Mistakes:
  • Assigning raw values instead of Node objects
  • Forgetting 'new' keyword for node creation
  • Confusing node value type requirements
5. You want to manually create this binary tree:
      8
     / \
    3   10
   / \    \
  1   6    14
     / \   /
    4   7 13

Which code snippet correctly creates this tree?
hard
A. Node root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(4); root.left.right.right = new Node(7); root.right.right = new Node(14); root.right.right.left = new Node(13);
B. Node root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(7); root.left.right.right = new Node(4); root.right.right = new Node(14); root.right.right.left = new Node(13);
C. Node root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(4); root.left.right.right = new Node(7); root.right.left = new Node(14); root.right.left.left = new Node(13);
D. Node root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(4); root.left.right.right = new Node(7); root.right.right = new Node(13); root.right.right.left = new Node(14);

Solution

  1. Step 1: Map tree structure to code

    Root 8 has left 3 and right 10. Node 3 has children 1 and 6. Node 6 has children 4 and 7. Node 10 has right child 14, which has left child 13.
  2. Step 2: Verify code matches tree

    Node root = new Node(8); root.left = new Node(3); root.right = new Node(10); root.left.left = new Node(1); root.left.right = new Node(6); root.left.right.left = new Node(4); root.left.right.right = new Node(7); root.right.right = new Node(14); root.right.right.left = new Node(13); correctly assigns all nodes to their proper left or right children matching the tree diagram.
  3. Final Answer:

    Option A code snippet correctly creates the tree -> Option A
  4. Quick Check:

    Match tree structure exactly in code [OK]
Hint: Match each node's left/right child exactly as tree diagram [OK]
Common Mistakes:
  • Swapping left and right children
  • Assigning children to wrong parent nodes
  • Mixing up node values in assignments
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