Inorder traversal gives sorted order in Data Structures Theory - Step-by-Step Execution

Concept Flow - Inorder traversal gives sorted order

Start at root

↓

Traverse left subtree

↓

Visit current node

↓

Traverse right subtree

↓

Repeat for each node

↓

All nodes visited in ascending order

Inorder traversal visits left subtree, then node, then right subtree, producing nodes in ascending order for binary search trees.

Execution Sample

Data Structures Theory

def inorder(node):
    if node is None:
        return
    inorder(node.left)
    print(node.value)
    inorder(node.right)

This code visits nodes in a binary search tree in ascending order using inorder traversal.

Analysis Table

Step	Current Node	Action	Output	Traversal Stack
1	10 (root)	Go left to 5		[10]
2	5	Go left to 2		[10, 5]
3	2	Go left to None		[10, 5, 2]
4	None	Back to 2, visit node	2	[10, 5]
5	2	Go right to None		[10, 5]
6	None	Back to 5, visit node	5	[10]
7	5	Go right to 7		[10, 7]
8	7	Go left to None		[10, 7]
9	None	Back to 7, visit node	7	[10]
10	7	Go right to None		[10]
11	None	Back to 10, visit node	10	[]
12	10	Go right to 15		[15]
13	15	Go left to None		[15]
14	None	Back to 15, visit node	15	[]
15	15	Go right to None		[]
16	None	Traversal complete		[]

💡 Traversal ends after visiting all nodes in ascending order.

State Tracker

Variable	Start	After Step 1	After Step 4	After Step 6	After Step 11	After Step 14	Final
Current Node	10	5	2	5	10	15	None
Output Sequence	[]	[]	[2]	[2,5]	[2,5,7,10]	[2,5,7,10,15]	[2,5,7,10,15]
Traversal Stack	[]	[10]	[10,5]	[10]	[]	[]	[]

Key Insights - 3 Insights

Why do we visit the current node only after traversing the left subtree?

Because inorder traversal follows left-node-right order, visiting left subtree first ensures smaller values come before the current node, producing sorted order as shown in steps 3 to 4 and 11.

What happens when the current node has no left child?

We immediately visit the current node since left subtree is empty, as seen at step 3 where going left leads to None, then back to visit node 2 at step 4.

How does the traversal stack help in inorder traversal?

The stack keeps track of nodes to return to after finishing left subtrees, enabling backtracking to visit nodes and then move right, as shown in the stack changes in the execution table.

Visual Quiz - 3 Questions

Look at the execution table at step 4. What node value is output?

A 2

B 5

C 7

D 10

💡 Hint

Check the 'Output' column at step 4 in the execution_table.

At which step does the traversal visit the root node (value 10)?

A Step 6

B Step 11

C Step 9

D Step 14

💡 Hint

Look for when 'Output' column shows 10 in the execution_table.

If the node with value 7 had a left child, how would the traversal stack change at step 8?

A It would remove 10 from the stack

B It would be empty

C It would include the new left child before 7

D It would not change

💡 Hint

Consider how the stack tracks nodes to visit left subtrees as shown in variable_tracker.

Concept Snapshot

Inorder traversal visits nodes in left-node-right order.
For binary search trees, this produces nodes in ascending sorted order.
Use recursion or stack to traverse left subtree, visit node, then right subtree.
Stack helps backtrack to parent nodes after left subtree is done.
Output sequence is sorted because left subtree nodes are smaller than current node.
Traversal ends after all nodes are visited once.

Full Transcript

Inorder traversal is a method to visit all nodes in a binary search tree in ascending order. It works by first visiting the left subtree, then the current node, and finally the right subtree. This order ensures that smaller values come before larger ones, producing a sorted sequence. The traversal uses a stack or recursion to remember nodes to return to after finishing left subtrees. The execution table shows step-by-step how the traversal moves left until it reaches a null child, then visits nodes and moves right. The variable tracker shows how the current node, output sequence, and stack change over time. Key moments clarify why the current node is visited after the left subtree and how the stack supports backtracking. The visual quiz tests understanding of output values at specific steps and how the stack changes if the tree structure changes. Overall, inorder traversal is a fundamental technique to get sorted order from binary search trees.

Practice

1. What is the main result of performing an inorder traversal on a binary search tree (BST)?

easy

A. It visits nodes randomly.

B. It visits nodes in descending sorted order.

C. It visits only leaf nodes.

D. It visits nodes in ascending sorted order.

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits the left child, then the node itself, then the right child.

2. Step 2: Apply inorder traversal to BST property

   Since BST nodes have smaller values on the left and larger on the right, inorder traversal visits nodes in ascending order.

3. Final Answer:

   It visits nodes in ascending sorted order. -> Option D

4. Quick Check:

   Inorder traversal = ascending order [OK]

Hint: Inorder on BST always gives sorted ascending list [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Thinking traversal visits nodes randomly
• Assuming it visits nodes in descending order

2. Which of the following is the correct sequence of steps in an inorder traversal of a binary tree?

easy

A. Visit left child, then node, then right child

B. Visit node, then left child, then right child

C. Visit right child, then node, then left child

D. Visit left child, then right child, then node

Solution

1. Step 1: Recall inorder traversal order

   Inorder traversal means visiting the left subtree first, then the node, then the right subtree.

2. Step 2: Match the correct sequence

   Visit left child, then node, then right child matches this order exactly: left child, node, right child.

3. Final Answer:

   Visit left child, then node, then right child -> Option A

4. Quick Check:

   Inorder = left, node, right [OK]

Hint: Inorder = left subtree, node, right subtree [OK]

Common Mistakes:

• Mixing up preorder and inorder sequences
• Visiting node before left child
• Visiting right child before node

3. Given the BST below, what is the output of an inorder traversal?

      10
     /  \
    5    15
   / \     \
  3   7     20

medium

A. [20, 15, 10, 7, 5, 3]

B. [10, 5, 3, 7, 15, 20]

C. [3, 5, 7, 10, 15, 20]

D. [5, 3, 7, 10, 15, 20]

Solution

1. Step 1: Perform inorder traversal on the BST

   Visit left subtree (3, 5, 7), then root (10), then right subtree (15, 20).

2. Step 2: Write nodes in visited order

   The order is 3, 5, 7, 10, 15, 20.

3. Final Answer:

   [3, 5, 7, 10, 15, 20] -> Option C

4. Quick Check:

   Inorder traversal output = sorted ascending list [OK]

Hint: Inorder traversal of BST = sorted ascending values [OK]

Common Mistakes:

• Listing nodes in preorder or postorder instead
• Confusing left and right subtree order
• Forgetting to include all nodes

4. A student wrote this inorder traversal code but it does not print nodes in sorted order for a BST. What is the mistake?

def inorder(node):
    if node:
        inorder(node.right)
        print(node.value)
        inorder(node.left)

medium

A. The function visits right child before left child, reversing order.

B. The function misses printing the node value.

C. The function does not check if node is None.

D. The function should print before visiting children.

Solution

1. Step 1: Analyze the traversal order in code

   The code visits right child first, then node, then left child, which is reverse of inorder.

2. Step 2: Understand effect on BST traversal

   This reversed order prints nodes in descending order, not ascending sorted order.

3. Final Answer:

   The function visits right child before left child, reversing order. -> Option A

4. Quick Check:

   Inorder must visit left before right [OK]

Hint: Inorder visits left child before right child [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Printing node value in wrong place
• Not checking for null node

5. You have a binary tree that is not a BST. You perform an inorder traversal and get the sequence [4, 2, 5, 1, 6]. Which of the following is true?

hard

A. The tree is a BST because inorder traversal is sorted.

B. The tree is not a BST because inorder traversal is not sorted.

C. Inorder traversal always gives sorted order regardless of tree type.

D. The tree must be balanced to get this sequence.

Solution

1. Step 1: Check if inorder traversal sequence is sorted

   The sequence [4, 2, 5, 1, 6] is not in ascending order.

2. Step 2: Understand BST property and inorder traversal

   In a BST, inorder traversal always produces a sorted ascending sequence. Since this is not sorted, the tree is not a BST.

3. Final Answer:

   The tree is not a BST because inorder traversal is not sorted. -> Option B

4. Quick Check:

   Inorder sorted = BST; not sorted = not BST [OK]

Hint: Inorder sorted? Yes BST; No not BST [OK]

Common Mistakes:

• Assuming inorder always sorts any tree
• Confusing balanced tree with BST
• Ignoring order of traversal output