Tree Traversal Inorder Left Root Right in DSA C++

Mental Model

Visit the left side first, then the current node, then the right side. This order visits nodes in a sorted way for binary search trees.

Analogy: Imagine reading a book where you first read the left page, then the middle page, then the right page, so you don't miss anything and keep the order.

    2
   / \
  1   3

Traversal order: 1 -> 2 -> 3

Dry Run Walkthrough

Input: Binary tree: root=2, left child=1, right child=3; perform inorder traversal

Goal: Print nodes in left-root-right order to get sorted output

Step 1: Start at root (2), move to left child (1)

Current node: 1
Tree: 2 -> left: 1, right: 3

Why: In inorder, we visit left subtree first

Step 2: Node 1 has no left child, print 1

Output: 1
Current node: 1

Why: Left subtree done, now visit root

Step 3: Move back to parent (2), print 2

Output: 1 2
Current node: 2

Why: After left subtree, visit root

Step 4: Move to right child (3), no left child, print 3

Output: 1 2 3
Current node: 3

Why: Visit right subtree after root

Result:

Output: 1 2 3

Annotated Code

DSA C++

#include <iostream>
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

void inorderTraversal(Node* root) {
    if (root == nullptr) return; // base case: empty node
    inorderTraversal(root->left); // visit left subtree
    cout << root->data << " "; // visit root
    inorderTraversal(root->right); // visit right subtree
}

int main() {
    Node* root = new Node(2);
    root->left = new Node(1);
    root->right = new Node(3);

    inorderTraversal(root);
    cout << endl;
    return 0;
}

if (root == nullptr) return; // base case: empty node

stop recursion when node is empty

inorderTraversal(root->left); // visit left subtree

recursively visit left child first

cout << root->data << " "; // visit root

print current node after left subtree

inorderTraversal(root->right); // visit right subtree

recursively visit right child last

Output

1 2 3 

Complexity Analysis

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

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Compared to iterative traversal, recursive is simpler but uses call stack; iterative uses explicit stack but same time complexity

Edge Cases

Empty tree (root is nullptr)

No output, function returns immediately

DSA C++

if (root == nullptr) return; // base case: empty node

Single node tree

Prints the single node value

DSA C++

if (root == nullptr) return; // base case: empty node

Left skewed tree (all nodes only have left child)

Prints nodes from bottom left up to root in order

DSA C++

inorderTraversal(root->left); // visit left subtree

When to Use This Pattern

When a problem asks to visit all nodes in sorted order in a binary search tree, use inorder traversal because it visits left, root, then right nodes.

Common Mistakes

Mistake: Visiting root before left subtree (preorder) instead of after
Fix: Move the print statement after the recursive call to left child

Mistake: Skipping the right subtree traversal
Fix: Add recursive call to right child after printing root

Summary

Visits nodes in left-root-right order to get sorted output for binary trees.

Use when you want to process nodes in ascending order or sorted sequence.

Remember: always visit left subtree first, then root, then right subtree.

Practice

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

easy

A. Right subtree, Root, Left subtree

B. Root, Left subtree, Right subtree

C. Left subtree, Root, Right subtree

D. Root, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left child, then root, then right child.

2. Step 2: Apply to binary tree traversal

   This means we first visit all nodes in the left subtree, then the root node, then all nodes in the right subtree.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option C

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder means Left, Root, then Right [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 C++ function signature for inorder traversal of a binary tree node pointer Node* root?

easy

A. int inorderTraversal(Node* root)

B. int inorderTraversal(Node root)

C. void inorderTraversal(Node root)

D. void inorderTraversal(Node* root)

Solution

1. Step 1: Identify correct parameter type

   Traversal functions usually take a pointer to the root node to access the tree nodes.

2. Step 2: Identify correct return type

   Traversal functions typically return void because they print or process nodes without returning a value.

3. Final Answer:

   void inorderTraversal(Node* root) -> Option D

4. Quick Check:

   Pointer parameter and void return [OK]

Hint: Traversal functions use Node* and void return [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using int return type instead of void
• Missing pointer symbol * in parameter

3. Given the binary tree:

    2
   / \
  1   3

What is the output of the inorder traversal function that prints node values separated by spaces?

medium

A. 1 2 3

B. 2 1 3

C. 3 2 1

D. 1 3 2

Solution

1. Step 1: Traverse left subtree

   Start at root (2), go left to node (1), print 1 since it has no left child.

2. Step 2: Visit root and right subtree

   Return to root (2), print 2, then go right to node (3), print 3.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Inorder traversal output = 1 2 3 [OK]

Hint: Inorder prints nodes in ascending order for BST [OK]

Common Mistakes:

• Printing root before left subtree
• Printing right subtree before root
• Confusing preorder or postorder output

4. Consider this C++ inorder traversal code snippet:

void inorder(Node* root) {
  if (root == nullptr) return;
  inorder(root->left);
  std::cout << root->val << " ";
  inorder(root->right);
}

What is the error if the output is empty for a non-empty tree?

medium

A. The base case condition is incorrect

B. The function is not called with the root node

C. The print statement is missing

D. The traversal order is wrong

Solution

1. Step 1: Check base case correctness

   The base case correctly returns when root is nullptr, so no error here.

2. Step 2: Verify function call

   If output is empty, likely the function was never called with the actual root node of the tree.

3. Final Answer:

   The function is not called with the root node -> Option B

4. Quick Check:

   Missing initial call causes no output [OK]

Hint: Always call traversal with actual root node [OK]

Common Mistakes:

• Not calling traversal function at all
• Calling with nullptr instead of root
• Assuming print statement missing when it exists

5. You have a binary tree with nodes:

      4
     / \
    2   6
   / \ / \
  1  3 5  7

Using inorder traversal, which sequence of node values will you get?

hard

A. [1, 2, 3, 4, 5, 6, 7]

B. [4, 2, 1, 3, 6, 5, 7]

C. [7, 6, 5, 4, 3, 2, 1]

D. [1, 3, 2, 5, 7, 6, 4]

Solution

1. Step 1: Traverse left subtree of root (4)

   Visit left subtree rooted at 2: inorder visits 1, then 2, then 3.

2. Step 2: Visit root and right subtree

   After left subtree, visit root 4, then right subtree rooted at 6: visit 5, then 6, then 7.

3. Final Answer:

   [1, 2, 3, 4, 5, 6, 7] -> Option A

4. Quick Check:

   Inorder traversal of BST gives sorted sequence [OK]

Hint: Inorder traversal of BST returns sorted node values [OK]

Common Mistakes:

• Mixing preorder or postorder sequences
• Visiting root before left subtree
• Visiting right subtree before root