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

Recall & Review

beginner

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

In Preorder traversal, nodes are visited in this order: Root first, then Left subtree, and finally Right subtree.

beginner

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

The Preorder traversal visits nodes as Root-Left-Right.
Output: 1 -> 2 -> 3

beginner

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

The base case is when the current node is nullptr. Then, the function returns without doing anything.

intermediate

Why is Preorder traversal useful in copying a tree?

Because Preorder visits the root before children, it allows copying the root node first, then recursively copying left and right subtrees, preserving structure.

beginner

Write the recursive steps for Preorder traversal in simple words.

1. Visit the current node (root).
2. Traverse the left subtree in Preorder.
3. Traverse the right subtree in Preorder.

In Preorder traversal, which node is visited first?

A Left child

B Right child

C Root node

D Leaf node

Preorder traversal always visits the root node first before its children.

What is the correct Preorder traversal sequence for this tree?
1
/ \
2 3

A 1 -> 2 -> 3

B 2 -> 1 -> 3

C 3 -> 2 -> 1

D 2 -> 3 -> 1

Preorder visits root first (1), then left child (2), then right child (3).

Which traversal order is Preorder?

A Left, Right, Root

B Root, Left, Right

C Left, Root, Right

D Right, Left, Root

Preorder traversal visits root first, then left subtree, then right subtree.

What happens when the recursive Preorder function reaches a null node?

A It continues to next node

B It visits the null node

C It throws an error

D It returns immediately

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

Which of these is NOT a use case of Preorder traversal?

A Printing nodes in sorted order

B Evaluating expression trees

C Saving tree structure

D Copying a tree

Printing nodes in sorted order requires Inorder traversal, not Preorder.

Explain how Preorder traversal visits nodes in a binary tree.

Think about the order Root, Left, Right.

Describe a simple recursive function for Preorder traversal.

Remember the base case and the three main steps.

Practice

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

easy

A. Right, Root, Left

B. Left, Root, Right

C. Left, Right, Root

D. Root, Left, Right

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then the left subtree, and finally the right subtree.

2. Step 2: Match the order with options

   Root, Left, Right matches the order Root, Left, Right exactly.

3. Final Answer:

   Root, Left, Right -> Option D

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Remember preorder starts with the root node first [OK]

Common Mistakes:

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

2. Which of the following C++ function signatures correctly represents a preorder traversal of a binary tree node Node*?

easy

A. void preorder(Node* root) { if(root) { cout << root->data; preorder(root->left); preorder(root->right); } }

B. void preorder(Node* root) { if(root) { preorder(root->left); preorder(root->right); cout << root->data; } }

C. void preorder(Node* root) { if(root) { preorder(root->right); cout << root->data; preorder(root->left); } }

D. void preorder(Node* root) { if(root) { preorder(root->left); cout << root->data; preorder(root->right); } }

Solution

1. Step 1: Recall preorder traversal order

   Preorder visits root first, then left subtree, then right subtree.

2. Step 2: Check function code order

   void preorder(Node* root) { if(root) { cout << root->data; preorder(root->left); preorder(root->right); } } prints root data first, then recursively visits left and right children, matching preorder.

3. Final Answer:

   void preorder(Node* root) { if(root) { cout << root->data; preorder(root->left); preorder(root->right); } } -> Option A

4. Quick Check:

   Print root first, then left, then right [OK]

Hint: Print root data before recursive calls for preorder [OK]

Common Mistakes:

• Printing after recursive calls (postorder)
• Swapping left and right recursive calls
• Missing null check for root

3. Given the binary tree below, what is the preorder traversal output?

      10
     /  \
    5    15
   /    /  \
  3    12  20

medium

A. [10, 15, 20, 12, 5, 3]

B. [10, 5, 3, 15, 12, 20]

C. [3, 5, 10, 12, 15, 20]

D. [5, 3, 10, 12, 15, 20]

Solution

1. Step 1: Start at root and visit nodes in preorder

   Visit root 10, then left subtree (5, then 3), then right subtree (15, then 12 and 20).

2. Step 2: Write traversal sequence

   Sequence is 10, 5, 3, 15, 12, 20.

3. Final Answer:

   [10, 5, 3, 15, 12, 20] -> Option B

4. Quick Check:

   Preorder = Root, Left, Right = [10,5,3,15,12,20] [OK]

Hint: Follow root, then left subtree, then right subtree order [OK]

Common Mistakes:

• Mixing preorder with inorder or postorder
• Visiting right subtree before left
• Skipping leaf nodes

4. Identify the error in the following preorder traversal function and select the corrected version:

void preorder(Node* root) {
  if(root == nullptr) return;
  preorder(root->left);
  cout << root->data << " ";
  preorder(root->right);
}

medium

A. void preorder(Node* root) { if(root == nullptr) return; cout << root->data << " "; preorder(root->left); preorder(root->right); }

B. void preorder(Node* root) { if(root == nullptr) return; preorder(root->right); cout << root->data << " "; preorder(root->left); }

C. void preorder(Node* root) { if(root == nullptr) return; preorder(root->left); preorder(root->right); cout << root->data << " "; }

D. void preorder(Node* root) { if(root == nullptr) return; cout << root->data << " "; preorder(root->right); preorder(root->left); }

Solution

1. Step 1: Identify preorder traversal order

   Preorder visits root first, then left subtree, then right subtree.

2. Step 2: Check given function and options

   The given function prints after left subtree, which is inorder traversal. void preorder(Node* root) { if(root == nullptr) return; cout << root->data << " "; preorder(root->left); preorder(root->right); } prints root first, then left, then right, matching preorder.

3. Final Answer:

   void preorder(Node* root) { if(root == nullptr) return; cout << root->data << " "; preorder(root->left); preorder(root->right); } -> Option A

4. Quick Check:

   Print root before recursive calls for preorder [OK]

Hint: Print root data before recursive calls to fix preorder [OK]

Common Mistakes:

• Printing after left subtree (inorder)
• Swapping left and right subtree calls
• Missing null pointer check

5. You have a binary tree where some nodes have only one child. How does preorder traversal help in serializing this tree uniquely, and which sequence below represents the preorder traversal of this tree?

       8
      / \
     4   10
      \    \
       6    20

hard

A. [8, 10, 20, 4, 6]

B. [4, 6, 8, 10, 20]

C. [8, 4, 6, 10, 20]

D. [6, 4, 8, 20, 10]

Solution

1. Step 1: Understand preorder traversal for serialization

   Preorder visits root first, then left subtree, then right subtree, capturing structure including nodes with single children.

2. Step 2: Traverse given tree in preorder

   Visit 8 (root), then left subtree: 4, then 6 (right child of 4), then right subtree: 10, then 20 (right child of 10).

3. Step 3: Write traversal sequence

   Sequence is [8, 4, 6, 10, 20].

4. Final Answer:

   [8, 4, 6, 10, 20] -> Option C

5. Quick Check:

   Preorder captures root-left-right including single children [OK]

Hint: Preorder captures tree shape by visiting root before children [OK]

Common Mistakes:

• Ignoring nodes with single child
• Mixing preorder with inorder
• Visiting right subtree before left