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Binary Tree Inorder Traversal

Easy
DFS

Given the root of a binary tree, return the inorder traversal of its nodes' values.

In an inorder traversal, we visit nodes in the following order:

  1. Recursively traverse the left subtree
  2. Visit the root node
  3. Recursively traverse the right subtree

For a Binary Search Tree (BST), inorder traversal returns nodes in sorted order.

Example Tree:

    1
     \
      2
     /
    3

Inorder traversal: [1, 3, 2]

Follow-up: Can you solve this iteratively using a stack?

Example 1

Input: root = [1,null,2,3]
Output: [1,3,2]
Explanation: Visit left (none), root (1), then right subtree. Right subtree: left (3), root (2), right (none).

Example 2

Input: root = []
Output: []
Explanation: Empty tree returns empty list.

Example 3

Input: root = [1]
Output: [1]
Explanation: Single node tree returns just that node.

Constraints

  • •The number of nodes in the tree is in the range [0, 100].
  • •-100 <= Node.val <= 100
Show Hints (4)
Hint 1: For recursive approach, what is the base case when the node is null?
Hint 2: The key insight is: process left subtree completely, then current node, then right subtree.
Hint 3: For iterative approach, use a stack to simulate the call stack. Go left as far as possible, then pop and go right.
Hint 4: Think about what you need to track: the current node and whether you've visited its left subtree.
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