Tree Traversal Preorder Root Left Right in DSA Javascript

Mental Model

Visit the root node first, then explore the left subtree, and finally the right subtree.

Analogy: Imagine reading a family tree starting from the oldest ancestor, then visiting their left child and all descendants before moving to the right child and their descendants.

    1
   / \
  2   3
 / \
4   5

↑ Root at 1

Dry Run Walkthrough

Input: Tree: 1 as root, 2 as left child, 3 as right child, 4 as left child of 2, 5 as right child of 2

Goal: Print nodes in preorder: root first, then left subtree, then right subtree

Step 1: Visit root node 1

Visited: 1

Why: Start traversal at root

Step 2: Traverse left subtree starting at node 2

Visited: 1 -> 2

Why: After root, visit left subtree

Step 3: Traverse left subtree of node 2, visit node 4

Visited: 1 -> 2 -> 4

Why: Preorder visits left child before right

Step 4: Back to node 2, traverse right subtree, visit node 5

Visited: 1 -> 2 -> 4 -> 5

Why: Complete left subtree before right subtree

Step 5: Traverse right subtree starting at node 3

Visited: 1 -> 2 -> 4 -> 5 -> 3

Why: After left subtree, visit right subtree

Result:

1 -> 2 -> 4 -> 5 -> 3

Annotated Code

DSA Javascript

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function preorderTraversal(root) {
  const result = [];
  function traverse(node) {
    if (node === null) return; // base case: no node
    result.push(node.val); // visit root
    traverse(node.left); // traverse left subtree
    traverse(node.right); // traverse right subtree
  }
  traverse(root);
  return result.join(' -> ');
}

// Build tree from dry_run input
const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);

console.log(preorderTraversal(root));

if (node === null) return; // base case: no node

stop traversal when no node to visit

result.push(node.val); // visit root

record current node value before children

traverse(node.left); // traverse left subtree

explore left subtree first

traverse(node.right); // traverse right subtree

explore right subtree after left

Output

1 -> 2 -> 4 -> 5 -> 3

Complexity Analysis

Time: O(n) because each node is visited exactly once

Space: O(n) due to recursion stack and result storage in worst case

vs Alternative: Compared to inorder or postorder, preorder visits nodes in a different order but has the same time and space complexity

Edge Cases

Empty tree (root is null)

Returns empty string, no nodes visited

DSA Javascript

if (node === null) return; // base case: no node

Tree with single node

Returns that single node value

DSA Javascript

if (node === null) return; // base case: no node

When to Use This Pattern

When you need to process or print nodes starting from the root before children, preorder traversal is the pattern to use.

Common Mistakes

Mistake: Visiting left or right child before the root node
Fix: Always visit the current node first before traversing children

Mistake: Not handling null nodes causing errors
Fix: Add base case to return immediately when node is null

Summary

Visits nodes starting from root, then left subtree, then right subtree

Use when you want to process root before its children, such as copying a tree

Remember: root first, then left, then right

Practice

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

easy

A. Left, Right, Root

B. Left, Root, Right

C. Root, Left, Right

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: Confirm the visiting order

   The visiting order is Root, then Left child, then Right child.

3. Final Answer:

   Root, Left, Right -> Option C

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 or right child before root
• Mixing up left and right subtree order

2. Which of the following is the correct JavaScript function signature for preorder traversal of a binary tree node named root?

easy

A. function preorder(root) { if(root) { preorder(root.left); preorder(root.right); console.log(root.value); } }

B. function preorder(root) { if(root) { preorder(root.left); console.log(root.value); preorder(root.right); } }

C. function preorder(root) { if(root) { preorder(root.right); preorder(root.left); console.log(root.value); } }

D. function preorder(root) { if(root) { console.log(root.value); preorder(root.left); 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 steps

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } prints root value first, then recursively visits left and right children, matching preorder.

3. Final Answer:

   function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } -> Option D

4. Quick Check:

   Print root before children = function preorder(root) { if(root) { console.log(root.value); preorder(root.left); preorder(root.right); } } [OK]

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

Common Mistakes:

• Printing after visiting children (postorder)
• Swapping left and right recursive calls
• Missing base case for null root

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

    1
   / \
  2   3
 / \
4   5

Use the format: 1 -> 2 -> 4 -> 5 -> 3 -> null

medium

A. 1 -> 2 -> 4 -> 5 -> 3 -> null

B. 4 -> 2 -> 5 -> 1 -> 3 -> null

C. 2 -> 4 -> 5 -> 1 -> 3 -> null

D. 1 -> 3 -> 2 -> 4 -> 5 -> null

Solution

1. Step 1: Start at root node 1

   Visit root first: 1

2. Step 2: Traverse left subtree of 1

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

3. Step 3: Traverse right subtree of 1

   Visit 3

4. Final Answer:

   1 -> 2 -> 4 -> 5 -> 3 -> null -> Option A

5. Quick Check:

   Preorder visits root, left, right = 1 -> 2 -> 4 -> 5 -> 3 -> null [OK]

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

Common Mistakes:

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

4. The following JavaScript code is intended to perform a preorder traversal and print node values. What is the error?

function preorder(root) {
  if (root === null) return;
  preorder(root.left);
  console.log(root.value);
  preorder(root.right);
}

medium

A. The console.log should be before the recursive calls

B. The base case should check for undefined, not null

C. The function should return a value instead of void

D. The left and right recursive calls are swapped

Solution

1. Step 1: Analyze the order of operations

   The code visits left child first, then prints root, then visits right child.

2. Step 2: Compare with preorder traversal order

   Preorder requires printing root before visiting children, so console.log should be before recursive calls.

3. Final Answer:

   The console.log should be before the recursive calls -> Option A

4. Quick Check:

   Print root first in preorder [OK]

Hint: Print root value before recursive calls in preorder [OK]

Common Mistakes:

• Printing after left subtree instead of before
• Confusing preorder with inorder traversal
• Ignoring base case for null nodes

5. You have a binary tree where some nodes have only one child. How does preorder traversal handle nodes with a missing left child?
Consider this tree:

    10
      \
      20
     /  \
   30    40

What is the preorder traversal output?

hard

A. 10 -> 40 -> 20 -> 30 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 10 -> 20 -> 40 -> 30 -> null

D. 10 -> 30 -> 20 -> 40 -> null

Solution

1. Step 1: Visit root node 10

   Start with root: 10

2. Step 2: Check left child of 10

   Left child is missing, so skip to right child 20

3. Step 3: Visit node 20 and its children

   Visit 20, then left child 30, then right child 40

4. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option B

5. Quick Check:

   Preorder visits root, left (if any), right = 10 -> 20 -> 30 -> 40 -> null [OK]

Hint: Skip missing children, visit existing nodes in preorder order [OK]

Common Mistakes:

• Assuming missing left child means skip right child
• Visiting children in wrong order
• Confusing preorder with other traversals