Tree Traversal Preorder Root Left Right in DSA C - Time & Space Complexity

Time Complexity: Tree Traversal Preorder Root Left Right

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

How does the number of steps change when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void preorderTraversal(Node* root) {
    if (root == NULL) return;
    printf("%d ", root->data);  // Visit root
    preorderTraversal(root->left);  // Traverse left subtree
    preorderTraversal(root->right); // Traverse right subtree
}
    

This code visits every node in the tree starting from the root, then left child, then right child.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once (printing data).
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

Each node is visited once, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Preorder traversal takes longer because it visits the root first and then children."

[OK] Correct: The order of visiting nodes does not change the total number of nodes visited; every node is still visited once.

Interview Connect

Understanding preorder traversal time helps you explain how tree algorithms scale and shows you can analyze recursive code clearly.

Self-Check

"What if we changed preorder traversal to inorder traversal? How would the time complexity change?"

Practice

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

easy

A. Root, Left, Right

B. Left, Root, Right

C. Left, Right, Root

D. Right, Root, Left

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 as defined.

3. Final Answer:

   Root, Left, Right -> Option A

4. Quick Check:

   Preorder = Root, Left, Right [OK]

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

Common Mistakes:

• Confusing preorder with inorder or postorder
• Swapping left and right subtree order
• Thinking traversal starts with left child

2. Which of the following is the correct C syntax to start a preorder traversal function for a binary tree node named root?

easy

A. void preorder(Node root*) { if(root == NULL) return; ... }

B. void preorder(Node root) { if(root == NULL) return; ... }

C. void preorder(Node *root) { if(root == NULL) return; ... }

D. void preorder(Node *root) { if(root = NULL) return; ... }

Solution

1. Step 1: Check function parameter type

   Binary tree nodes are usually passed as pointers, so Node *root is correct.

2. Step 2: Check null pointer check syntax

   The condition if(root == NULL) correctly checks for null pointer. Assignment = is wrong.

3. Final Answer:

   void preorder(Node *root) { if(root == NULL) return; ... } -> Option C

4. Quick Check:

   Pointer param + null check == void preorder(Node *root) { if(root == NULL) return; ... } [OK]

Hint: Use pointer and '==' for null check in C [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using assignment '=' instead of comparison '=='
• Incorrect pointer syntax like 'Node root*'

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

    1
   / \
  2   3
 / \
4   5

medium

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

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

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

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

Solution

1. Step 1: Start at root node 1

   Preorder visits root first: 1

2. Step 2: Traverse left subtree of 1

   Visit 2, then left child 4, then right child 5: 2, 4, 5

3. Step 3: Traverse right subtree of 1

   Visit right child 3: 3

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

• Mixing preorder with inorder or postorder
• Visiting children before root
• Swapping left and right subtree order

4. Identify the error in this preorder traversal function and select the fix:

void preorder(Node *root) {
  if(root == NULL) return;
  preorder(root->left);
  preorder(root->right);
  printf("%d ", root->data);
}

medium

A. Add return type 'int' instead of 'void'

B. Change 'root == NULL' to 'root = NULL' in if condition

C. Remove recursive call to right subtree

D. Move printf before recursive calls to print root first

Solution

1. Step 1: Analyze traversal order in code

   Current code prints after left and right calls, which is postorder (Left, Right, Root).

2. Step 2: Fix order to preorder

   Print root data before recursive calls to get Root, Left, Right order.

3. Final Answer:

   Move printf before recursive calls to print root first -> Option D

4. Quick Check:

   Preorder prints root before children [OK]

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

Common Mistakes:

• Printing after recursive calls (postorder)
• Using assignment '=' instead of comparison '=='
• Changing return type unnecessarily

5. You want to collect all node values in preorder traversal into an array instead of printing. Which approach correctly modifies the preorder function in C?

void preorder(Node *root, int *arr, int *index) {
  if(root == NULL) return;
  // What goes here?
}

hard

A. arr[*index] = root->data; (*index)++; preorder(root->right, arr, index); preorder(root->left, arr, index);

B. arr[(*index)++] = root->data; preorder(root->left, arr, index); preorder(root->right, arr, index);

C. preorder(root->left, arr, index); preorder(root->right, arr, index); arr[(*index)++] = root->data;

D. preorder(root->left, arr, index); arr[(*index)++] = root->data; preorder(root->right, arr, index);

Solution

1. Step 1: Understand preorder traversal order

   Preorder visits root first, so store root data before recursive calls.

2. Step 2: Correctly update array and index

   Use arr[(*index)++] = root->data; to store and increment index, then recurse left and right.

3. Final Answer:

   arr[(*index)++] = root->data; preorder(root->left, arr, index); preorder(root->right, arr, index); -> Option B

4. Quick Check:

   Store root first, then left and right [OK]

Hint: Store root data before recursive calls, increment index [OK]

Common Mistakes:

• Storing data after recursive calls (not preorder)
• Incrementing index incorrectly
• Swapping left and right recursive calls