What is the brute force approach and complexity for Binary Tree Inorder Traversal?

Recurse into the left subtree, record the node, then recurse into the right subtree with time complexity O(N) and space complexity O(H).

What is the optimal approach and complexity for Binary Tree Inorder Traversal?

Use empty right pointers as links back up the tree; no stack, O(1) extra space with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Binary Tree Inorder Traversal?

Look for key signals in the prompt: "inorder traversal" indicates Left side, then the node, then the right side. The node waits for its left side, which is the whole point of this question.; "return the list of values" indicates We need the values in order. The output list is not counted as extra space.; "follow-up: iterative" indicates The recursion's call stack has to become a stack we manage (Basic) or threads (Morris)..

Binary Tree Inorder Traversal

Easy·LeetCode #94 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given the root of a binary tree. Return the values of all its nodes in inorder: first everything in the left subtree, then the node itself, then everything in the right subtree.

The recursive version is short. The follow-up asks you to write it iteratively, without recursion.

Input

The root of a binary tree. The tree may be empty.

Output

A list of the node values, in inorder.

Example 1

Input: root = [1, null, 2, 3]

Output: [1, 3, 2]

Node 1 has no left child, so 1 comes first. Node 2 has the left child 3, so 3 comes before 2.

Example 2

Input: root = [1, 2, 3, 4, 5, null, 8, null, null, 6, 7, 9]

Output: [4, 2, 6, 5, 7, 1, 3, 9, 8]

The left side of 1 is 2, with children 4 and 5 (and 5 has children 6 and 7). It gives 4, 2, 6, 5, 7. Then comes 1. The right side is 3 with no left child, and its right child 8 has the left child 9. It gives 3, 9, 8.

Example 3

Input: root = []

Output: []

There are no nodes, so the list is empty.

Constraints

• The number of nodes in the tree is in the range [0, 100].
• −100 ≤ Node.val ≤ 100

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“inorder traversal”

→ Left side, then the node, then the right side. The node waits for its left side, which is the whole point of this question.

“return the list of values”

→ We need the values in order. The output list is not counted as extra space.

“follow-up: iterative”

→ The recursion's call stack has to become a stack we manage (Basic) or threads (Morris).

Approach Complexity Ladder

Brute Force — recursionO(N)O(H)

Recurse into the left subtree, record the node, then recurse into the right subtree

Better — iterative with an explicit stackO(N)O(H)

Do the recursion by hand: walk left and push, pop to visit, then walk right

Optimal — Morris traversal (threads)O(N)O(1)

Use empty right pointers as links back up the tree; no stack, O(1) extra space

Brute Force Intuition

Inorder is already a recursive rule: the left part, then this node, then the right part. Writing the rule as a function call is the whole solution. The call stack does the bookkeeping, which is why it is short.

Recognition rule for this sub-pattern:

Visit nodes in pre/in/post order or walk two trees together node by node.

Remember this · Tree DFS

Trigger

Paths, heights, diameters, validating structure, ancestors, anything "for every subtree".

Think

If my left and right children already gave me their answers, how do I build mine?

Key question

What does my recursive function return, and what do I pass down?

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