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Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.37

94\. Binary Tree Inorder Traversal

Easy

Given the `root` of a binary tree, return _the inorder traversal of its nodes' values_.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg)

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

**Output:** [1,3,2] 

**Example 2:**

**Input:** root = []

**Output:** [] 

**Example 3:**

**Input:** root = [1]

**Output:** [1] 

**Example 4:**

![](https://assets.leetcode.com/uploads/2020/09/15/inorder_5.jpg)

**Input:** root = [1,2]

**Output:** [2,1] 

**Example 5:**

![](https://assets.leetcode.com/uploads/2020/09/15/inorder_4.jpg)

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

**Output:** [1,2] 

**Constraints:**

*   The number of nodes in the tree is in the range `[0, 100]`.
*   `-100 <= Node.val <= 100`

**Follow up:** Recursive solution is trivial, could you do it iteratively?

To solve the "Binary Tree Inorder Traversal" problem in Java with the Solution class, follow these steps:

1. Define a method `inorderTraversal` in the `Solution` class that takes the root of a binary tree as input and returns the inorder traversal of its nodes' values.
2. Implement an iterative algorithm to perform inorder traversal:
   - Initialize an empty list to store the inorder traversal result.
   - Initialize a stack to track the nodes during traversal.
   - Start with the root node and push it onto the stack.
   - While the stack is not empty:
     - Traverse down the left subtree by pushing all left child nodes onto the stack.
     - Pop the top node from the stack and add its value to the traversal result list.
     - Move to the right subtree of the popped node and repeat the process.
   - Return the traversal result list.
3. Return the inorder traversal result list.

Here's the implementation of the `inorderTraversal` method in Java:

```java
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

class Solution {
    public List inorderTraversal(TreeNode root) {
        List inorder = new ArrayList<>();
        Stack stack = new Stack<>();
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            inorder.add(curr.val);
            curr = curr.right;
        }
        return inorder;
    }
}
```

This implementation performs an iterative inorder traversal of the binary tree using a stack, with a time complexity of O(N), where N is the number of nodes in the tree.