Tree Traversal Inorder Left Root Right in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the order of nodes visited in an inorder tree traversal?

In inorder traversal, nodes are visited in the order: Left subtree, then Root node, and finally Right subtree.

beginner

Why is inorder traversal useful for binary search trees (BST)?

Inorder traversal of a BST visits nodes in ascending sorted order because it visits left (smaller), root, then right (larger) nodes.

beginner

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

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

beginner

Show the inorder traversal output for this tree:
Root=2, Left=1, Right=3

The inorder traversal visits nodes in order: 1 (left), 2 (root), 3 (right). So output is: 1 -> 2 -> 3

beginner

What is the time complexity of inorder traversal on a binary tree with n nodes?

The time complexity is O(n) because each node is visited exactly once.

In inorder traversal, which node is visited first?

A Left child

B Root node

C Right child

D None

In inorder traversal, the left subtree is visited first before the root.

What is the output of inorder traversal on a BST with nodes 5, 3, 7?

A 5 -> 3 -> 7

B 3 -> 5 -> 7

C 7 -> 5 -> 3

D 3 -> 7 -> 5

Inorder traversal visits left, root, right, so nodes are visited in ascending order.

What happens when the inorder traversal function reaches a null node?

A It visits the null node

B It throws an error

C It stops and returns

D It skips to the right subtree

The base case for recursion is when the node is null, so the function returns immediately.

Which traversal order is correct for inorder?

A Root, Left, Right

B Root, Right, Left

C Right, Root, Left

D Left, Root, Right

Inorder traversal always visits left subtree first, then root, then right subtree.

How many times is each node visited in inorder traversal?

A Once

B Twice

C Three times

D Depends on tree height

Each node is visited exactly once during inorder traversal.

Explain the steps of inorder traversal on a binary tree.

Think about visiting left first, then root, then right.

Describe why inorder traversal outputs sorted nodes in a binary search tree.

Connect BST ordering with traversal order.

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Right subtree, Root, Left subtree

B. Root, Left subtree, Right subtree

C. Left subtree, Root, Right subtree

D. Root, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left child, then root, then right child.

2. Step 2: Apply to binary tree traversal

   This means we first visit all nodes in the left subtree, then the root node, then all nodes in the right subtree.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option C

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder means Left, Root, then Right [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Visiting root before left subtree
• Visiting right subtree before root

2. Which of the following is the correct C++ function signature for inorder traversal of a binary tree node pointer Node* root?

easy

A. int inorderTraversal(Node* root)

B. int inorderTraversal(Node root)

C. void inorderTraversal(Node root)

D. void inorderTraversal(Node* root)

Solution

1. Step 1: Identify correct parameter type

   Traversal functions usually take a pointer to the root node to access the tree nodes.

2. Step 2: Identify correct return type

   Traversal functions typically return void because they print or process nodes without returning a value.

3. Final Answer:

   void inorderTraversal(Node* root) -> Option D

4. Quick Check:

   Pointer parameter and void return [OK]

Hint: Traversal functions use Node* and void return [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using int return type instead of void
• Missing pointer symbol * in parameter

3. Given the binary tree:

    2
   / \
  1   3

What is the output of the inorder traversal function that prints node values separated by spaces?

medium

A. 1 2 3

B. 2 1 3

C. 3 2 1

D. 1 3 2

Solution

1. Step 1: Traverse left subtree

   Start at root (2), go left to node (1), print 1 since it has no left child.

2. Step 2: Visit root and right subtree

   Return to root (2), print 2, then go right to node (3), print 3.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Inorder traversal output = 1 2 3 [OK]

Hint: Inorder prints nodes in ascending order for BST [OK]

Common Mistakes:

• Printing root before left subtree
• Printing right subtree before root
• Confusing preorder or postorder output

4. Consider this C++ inorder traversal code snippet:

void inorder(Node* root) {
  if (root == nullptr) return;
  inorder(root->left);
  std::cout << root->val << " ";
  inorder(root->right);
}

What is the error if the output is empty for a non-empty tree?

medium

A. The base case condition is incorrect

B. The function is not called with the root node

C. The print statement is missing

D. The traversal order is wrong

Solution

1. Step 1: Check base case correctness

   The base case correctly returns when root is nullptr, so no error here.

2. Step 2: Verify function call

   If output is empty, likely the function was never called with the actual root node of the tree.

3. Final Answer:

   The function is not called with the root node -> Option B

4. Quick Check:

   Missing initial call causes no output [OK]

Hint: Always call traversal with actual root node [OK]

Common Mistakes:

• Not calling traversal function at all
• Calling with nullptr instead of root
• Assuming print statement missing when it exists

5. You have a binary tree with nodes:

      4
     / \
    2   6
   / \ / \
  1  3 5  7

Using inorder traversal, which sequence of node values will you get?

hard

A. [1, 2, 3, 4, 5, 6, 7]

B. [4, 2, 1, 3, 6, 5, 7]

C. [7, 6, 5, 4, 3, 2, 1]

D. [1, 3, 2, 5, 7, 6, 4]

Solution

1. Step 1: Traverse left subtree of root (4)

   Visit left subtree rooted at 2: inorder visits 1, then 2, then 3.

2. Step 2: Visit root and right subtree

   After left subtree, visit root 4, then right subtree rooted at 6: visit 5, then 6, then 7.

3. Final Answer:

   [1, 2, 3, 4, 5, 6, 7] -> Option A

4. Quick Check:

   Inorder traversal of BST gives sorted sequence [OK]

Hint: Inorder traversal of BST returns sorted node values [OK]

Common Mistakes:

• Mixing preorder or postorder sequences
• Visiting root before left subtree
• Visiting right subtree before root