Tree Traversal Inorder Left Root Right in DSA Javascript - Build from Scratch

Tree Traversal Inorder Left Root Right

📖 Scenario: You have a simple family tree where each person can have a left child and a right child. You want to visit each person in a special order: first the left child, then the person themselves, and then the right child. This is called inorder traversal.

🎯 Goal: Build a program that creates a small tree, sets up a list to store the traversal result, writes the inorder traversal function, and finally prints the order in which people are visited.

📋 What You'll Learn

Create a tree node structure with value, left, and right properties

Build a tree with exactly these nodes and connections: root 'A', left child 'B', right child 'C', left child of 'B' is 'D', right child of 'B' is 'E'

Create an empty array called result to store traversal output

Write a function called inorderTraversal that visits nodes in left-root-right order and adds their values to result

Call inorderTraversal with the root node

Print the result array after traversal

💡 Why This Matters

🌍 Real World

Tree traversal is used in many real-world applications like searching family trees, organizing files, and parsing expressions.

💼 Career

Understanding tree traversal is important for software developers working with data structures, databases, and algorithms.

1

Create the tree nodes and build the tree

Create variables root, nodeB, nodeC, nodeD, and nodeE as objects with value, left, and right properties. Set their values to 'A', 'B', 'C', 'D', and 'E' respectively. Connect them so that root.left = nodeB, root.right = nodeC, nodeB.left = nodeD, and nodeB.right = nodeE. All other children should be null.

DSA Javascript

// Create tree nodes and connect them
// Your code here

Hint

Each node is an object with value, left, and right. Connect them by assigning the left and right properties.

2

Create an empty array to store traversal results

Create an empty array called result to hold the values of nodes visited during inorder traversal.

DSA Javascript

const nodeD = { value: 'D', left: null, right: null };
const nodeE = { value: 'E', left: null, right: null };
const nodeB = { value: 'B', left: null, right: null };
const nodeC = { value: 'C', left: null, right: null };
const root = { value: 'A', left: null, right: null };
root.left = nodeB;
root.right = nodeC;
nodeB.left = nodeD;
nodeB.right = nodeE;

// Create an empty array called result
// Your code here

Hint

Use const result = []; to create an empty array.

3

Write the inorder traversal function

Write a function called inorderTraversal that takes a node as input. If the node is not null, first call inorderTraversal on the node's left child, then add the node's value to the result array, and finally call inorderTraversal on the node's right child.

DSA Javascript

const nodeD = { value: 'D', left: null, right: null };
const nodeE = { value: 'E', left: null, right: null };
const nodeB = { value: 'B', left: null, right: null };
const nodeC = { value: 'C', left: null, right: null };
const root = { value: 'A', left: null, right: null };
root.left = nodeB;
root.right = nodeC;
nodeB.left = nodeD;
nodeB.right = nodeE;

const result = [];

// Write the inorderTraversal function here
// Your code here

Hint

Use a function with a check for null. Visit left child first, then add current node's value, then visit right child.

4

Call the traversal and print the result

Call inorderTraversal with the root node. Then print the result array using console.log(result).

DSA Javascript

const nodeD = { value: 'D', left: null, right: null };
const nodeE = { value: 'E', left: null, right: null };
const nodeB = { value: 'B', left: null, right: null };
const nodeC = { value: 'C', left: null, right: null };
const root = { value: 'A', left: null, right: null };
root.left = nodeB;
root.right = nodeC;
nodeB.left = nodeD;
nodeB.right = nodeE;

const result = [];

function inorderTraversal(node) {
    if (node !== null) {
        inorderTraversal(node.left);
        result.push(node.value);
        inorderTraversal(node.right);
    }
}

// Call inorderTraversal with root and print result
// Your code here

Hint

Call the function with root and then print result with console.log(result).

Practice

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

easy

A. Root, Right subtree, Left subtree

B. Left subtree, Root, Right subtree

C. Right subtree, Root, Left subtree

D. Root, Left subtree, Right subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left subtree first, then the root node, and finally the right subtree.

2. Step 2: Match the traversal order to options

   Left subtree, Root, Right subtree correctly states the order as Left subtree, Root, Right subtree.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option B

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Inorder always visits left child before root [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 JavaScript function signature for inorder traversal of a binary tree node named node?

easy

A. function inorder(node) { inorder(node.right); console.log(node.value); inorder(node.left); }

B. function inorder(node) { console.log(node.value); inorder(node.left); inorder(node.right); }

C. function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } }

D. function inorder(node) { inorder(node.left); inorder(node.right); console.log(node.value); }

Solution

1. Step 1: Recall inorder traversal steps in code

   Inorder traversal visits left subtree, then root, then right subtree. So the function must call inorder on left, print root, then call inorder on right.

2. Step 2: Check each option's order

   function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } matches the correct order: left, root (console.log), right. Others have wrong orders.

3. Final Answer:

   function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } -> Option C

4. Quick Check:

   Left, Root, Right in code = function inorder(node) { if(node) { inorder(node.left); console.log(node.value); inorder(node.right); } } [OK]

Hint: Inorder prints after left call, before right call [OK]

Common Mistakes:

• Printing before left subtree traversal
• Swapping left and right subtree calls
• Missing null check for node

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

    10
   /  \
  5    15
 /    /  \
3    12  20

medium

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

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

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

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

Solution

1. Step 1: Traverse left subtree of root 10

   Left subtree of 10 is rooted at 5. Its left child is 3. Inorder of left subtree: visit 3, then 5 (root of subtree).

2. Step 2: Visit root 10 and then right subtree

   After left subtree, visit root 10. Then traverse right subtree rooted at 15: left child 12, root 15, right child 20. So inorder right subtree is 12, 15, 20.

3. Final Answer:

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

4. Quick Check:

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

Hint: List nodes left-root-right recursively [OK]

Common Mistakes:

• Visiting root before left subtree
• Mixing order of right subtree nodes
• Skipping leaf nodes

4. The following JavaScript code is intended to perform inorder traversal but has a bug. What is the bug?

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

medium

A. It prints the root before left subtree, so not inorder

B. It does not check for null nodes

C. It misses traversing the right subtree

D. It calls inorder on left and right in wrong order

Solution

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

   The code prints node.value first, then calls inorder on left and right. This is preorder traversal order.

2. Step 2: Identify correct inorder order

   Inorder traversal requires visiting left subtree first, then printing root, then right subtree. The code prints root before left subtree, which is incorrect.

3. Final Answer:

   It prints the root before left subtree, so not inorder -> Option A

4. Quick Check:

   Print after left call for inorder [OK]

Hint: Print after left subtree call for inorder [OK]

Common Mistakes:

• Printing root before left subtree
• Confusing preorder with inorder
• Not checking for null nodes (though code does here)

5. You have a binary search tree (BST) and want to get all values in ascending order using inorder traversal. Which of the following modifications to the inorder traversal function will correctly collect values into an array instead of printing them?

function inorder(node, arr) {
  if (!node) return;
  // Fill missing code here
}

hard

A. inorder(node.left, arr); inorder(node.right, arr); arr.push(node.value);

B. arr.push(node.value); inorder(node.left, arr); inorder(node.right, arr);

C. inorder(node.right, arr); arr.push(node.value); inorder(node.left, arr);

D. inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr);

Solution

1. Step 1: Recall inorder traversal order for BST

   Inorder traversal visits left subtree, then root, then right subtree, which returns values in ascending order for BST.

2. Step 2: Choose code that collects values in correct order

   inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); calls inorder on left, then pushes current node value, then calls inorder on right, correctly collecting values in ascending order.

3. Final Answer:

   inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); -> Option D

4. Quick Check:

   Left, Root, Right with push = inorder(node.left, arr); arr.push(node.value); inorder(node.right, arr); [OK]

Hint: Push after left call, before right call for sorted array [OK]

Common Mistakes:

• Pushing before left subtree traversal
• Swapping left and right subtree calls
• Pushing after both subtree traversals