How do you solve Binary Tree Inorder Traversal on LeetCode?

Binary Tree Inorder Traversal (LeetCode #94) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Inorder traversal visits nodes in a specific order: left, root, right.

What is the brute force solution for Binary Tree Inorder Traversal?

The brute force approach for Binary Tree Inorder Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves using recursion to traverse the tree in an inorder manner. This means we visit the left subtree, then the current node, and finally the right subtree. While this is straightforward, it can be inefficient in terms of space. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Inorder Traversal?

Binary Tree Inorder Traversal uses the following concepts: Stack, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Stack, Depth-First Search.

What are the interview tips for Binary Tree Inorder Traversal?

Explain your thought process clearly while coding. Consider edge cases, like an empty tree or a tree with only one node. Practice both recursive and iterative solutions to be well-rounded.

What are common mistakes when solving Binary Tree Inorder Traversal?

Forgetting to handle null nodes in the iterative solution. Confusing the order of traversal (left-root-right) with other traversal methods.

#94

Binary Tree Inorder Traversal

Easy

StackTreeDepth-First SearchBinary TreeStackDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses an iterative approach with a stack to simulate the recursive behavior of inorder traversal. This allows us to avoid the overhead of recursive calls while maintaining the same order of traversal.

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Algorithm

5 steps

1Step 1: Initialize an empty stack and a current node pointer starting at the root.
2Step 2: While the current node is not null or the stack is not empty, do the following:
3Step 3: Push all left children of the current node onto the stack.
4Step 4: Pop the top node from the stack, add its value to the result list, and move to its right child.
5Step 5: Repeat until both the stack is empty and the current node is null.

solution.py10 lines

1def inorderTraversal(root):
2    result, stack, current = [], [], root
3    while current or stack:
4        while current:
5            stack.append(current)
6            current = current.left
7        current = stack.pop()
8        result.append(current.val)
9        current = current.right
10    return result

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Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the stack used to store nodes during traversal.

1Inorder traversal visits nodes in a specific order: left, root, right.
2Using a stack allows us to simulate the recursive call stack without actual recursion.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.