Tree Traversal Preorder Root Left Right in DSA Typescript - Execution Trace

Concept Flow - Tree Traversal Preorder Root Left Right

Start at Root Node

↓

Visit Node: Process Data

↓

Traverse Left Subtree

↓

Traverse Right Subtree

↓

Done

Start at the root, visit it, then recursively do the same for left subtree, then right subtree.

Execution Sample

DSA Typescript

function preorder(node) {
  if (!node) return;
  console.log(node.val);
  preorder(node.left);
  preorder(node.right);
}

Prints nodes in preorder: root first, then left subtree, then right subtree.

Execution Table

Step	Operation	Node Visited	Left/Right	Tree State (Visual)
1	Visit root	A	Root	A ├─ B └─ C
2	Traverse left subtree	B	Left of A	A ├─ B │ ├─ D │ └─ E └─ C
3	Traverse left subtree	D	Left of B	A ├─ B │ ├─ D │ └─ E └─ C
4	Left child null	null	Left of D	No change
5	Right child null	null	Right of D	No change
6	Traverse right subtree	E	Right of B	A ├─ B │ ├─ D │ └─ E └─ C
7	Left child null	null	Left of E	No change
8	Right child null	null	Right of E	No change
9	Traverse right subtree	C	Right of A	A ├─ B │ ├─ D │ └─ E └─ C └─ F
10	Left child null	null	Left of C	No change
11	Traverse right subtree	F	Right of C	A ├─ B │ ├─ D │ └─ E └─ C └─ F
12	Left child null	null	Left of F	No change
13	Right child null	null	Right of F	No change
14	Traversal complete	null	None	Traversal done

💡 All nodes visited in preorder: root, left subtree, right subtree.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 6	After Step 9	After Step 11	Final
Current Node	A	A	B	D	E	C	F	null
Call Stack	[]	[preorder(A)]	[preorder(B), preorder(A)]	[preorder(D), preorder(B), preorder(A)]	[preorder(E), preorder(B), preorder(A)]	[preorder(C), preorder(A)]	[preorder(F), preorder(C), preorder(A)]	[]
Output Sequence	[]	[A]	[A, B]	[A, B, D]	[A, B, D, E]	[A, B, D, E, C]	[A, B, D, E, C, F]	[A, B, D, E, C, F]

Key Moments - 3 Insights

Why do we print the node value before traversing left and right children?

Because preorder traversal visits the root node first, then left subtree, then right subtree, as shown in execution_table steps 1, 2, and 9.

What happens when a node has no left or right child?

The traversal calls preorder with null, which returns immediately without action, as seen in steps 4, 5, 7, 8, 10, 12, and 13.

How does the call stack change during traversal?

Each recursive call adds a function to the stack; when a node is fully processed, its call returns and is removed, shown in variable_tracker's Call Stack column.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, which node is currently visited?

A Node E

B Node B

C Node D

D Node C

💡 Hint

Check the 'Node Visited' column at step 3 in execution_table.

At which step does the traversal visit the right child of node A?

A Step 2

B Step 9

C Step 6

D Step 11

💡 Hint

Look for 'Traverse right subtree' operation with node C in execution_table.

If node B had no left child, which step would be skipped?

A Step 4

B Step 6

C Step 3

D Step 7

💡 Hint

Step 4 is visiting left child null of D; if no left child of B, step 4 would be skipped.

Concept Snapshot

Preorder Tree Traversal:
- Visit root node first
- Recursively traverse left subtree
- Recursively traverse right subtree
- Uses recursion and call stack
- Output order: Root, Left, Right

Full Transcript

Preorder traversal visits nodes starting from the root, then left subtree, then right subtree. The code recursively visits each node, printing its value before going deeper. When a node is null, the recursion returns immediately. The call stack grows with each recursive call and shrinks as calls return. The output sequence follows the order nodes are visited.

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, Left, Root

Solution

1. Step 1: Understand preorder traversal definition

   Preorder traversal visits the root node first, then recursively visits the left subtree, followed by the right subtree.

2. Step 2: Match the order with options

   The order Root, Left, Right matches the preorder traversal definition exactly.

3. Final Answer:

   Root, Left, Right -> Option A

4. Quick Check:

   Preorder = Root, Left, Right [OK]

Hint: Preorder always starts with the root node first [OK]

Common Mistakes:

• Confusing preorder with inorder or postorder
• Visiting left or right before root
• Mixing up traversal orders

2. Which of the following TypeScript function signatures correctly represents a preorder traversal function for a binary tree node?

easy

A. function preorder(node: TreeNode): number[]

B. function preorder(node: TreeNode | null): void

C. function preorder(): TreeNode

D. function preorder(node: TreeNode, result: number[]): number

Solution

1. Step 1: Analyze function parameters and return type

   A preorder traversal function typically takes a node (or null) and returns void if it prints or processes nodes directly.

2. Step 2: Check each option

   function preorder(node: TreeNode | null): void accepts a node or null and returns void, which fits a traversal that processes nodes without returning a value.

3. Final Answer:

   function preorder(node: TreeNode | null): void -> Option B

4. Quick Check:

   Traversal function signature with node or null and void return [OK]

Hint: Traversal functions usually take a node and return void or array [OK]

Common Mistakes:

• Missing null in node parameter type
• Wrong return type for traversal
• Incorrect number of parameters

3. Given the following binary tree:

    1
   / \
  2   3
 / \
4   5

What is the output of a preorder traversal (Root, Left, Right)?

medium

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

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

C. [1, 2, 4, 5, 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 node 2, then its left child 4, then right child 5: 2, 4, 5

3. Step 3: Traverse right subtree of 1

   Visit node 3: 3

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

• Visiting left subtree after right subtree
• Starting traversal from left child instead of root
• Mixing preorder with inorder or postorder

4. Consider this TypeScript preorder traversal function:

function preorder(node: TreeNode | null): void {
  if (node === null) return;
  preorder(node.left);
  console.log(node.val);
  preorder(node.right);
}

What is the error in this code?

medium

A. It does not handle null nodes

B. It will cause infinite recursion

C. It prints nodes in postorder

D. It visits nodes in inorder, not preorder

Solution

1. Step 1: Analyze the order of operations in the function

   The function first calls preorder on left child, then prints current node, then calls preorder on right child.

2. Step 2: Compare with preorder definition

   Preorder should print the node first, then traverse left and right. This code prints after left subtree, so it is inorder traversal.

3. Final Answer:

   It visits nodes in inorder, not preorder -> Option D

4. Quick Check:

   Print before recursive calls = preorder; here print is in middle [OK]

Hint: Print node value before recursive calls for preorder [OK]

Common Mistakes:

• Printing node after left subtree instead of before
• Confusing traversal orders
• Missing base case for null

5. You want to collect all node values of a binary tree in preorder traversal into an array. Which TypeScript function correctly implements this?

hard

A. function preorder(node: TreeNode | null, result: number[] = []): number[] {
if (!node) return result;
result.push(node.val);
preorder(node.left, result);
preorder(node.right, result);
return result;
}

B. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [...preorder(node.left), node.val, ...preorder(node.right)];
}

C. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [...preorder(node.left), ...preorder(node.right), node.val];
}

D. function preorder(node: TreeNode | null): number[] {
if (!node) return [];
return [node.val, ...preorder(node.right), ...preorder(node.left)];
}

Solution

1. Step 1: Understand preorder traversal collection

   Preorder visits root first, then left subtree, then right subtree, collecting values in order.

2. Step 2: Check each function's order and correctness

   function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } pushes node.val first, then recursively collects left and right, returning the result array correctly.

3. Final Answer:

   function preorder(node: TreeNode | null, result: number[] = []): number[] { if (!node) return result; result.push(node.val); preorder(node.left, result); preorder(node.right, result); return result; } -> Option A

4. Quick Check:

   Push root first, then left and right recursively [OK]

Hint: Push root value before recursive calls to left and right [OK]

Common Mistakes:

• Collecting left subtree before root
• Mixing left and right subtree order
• Returning new arrays instead of accumulating