> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/0094-binary-tree-inorder-traversal.md).

# 0094. Binary Tree Inorder Traversal

<https://leetcode.com/problems/binary-tree-inorder-traversal>

## Description

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?

## AC1: Divide and conquer

Like a queen, order two worker bee to report their result, and then combine her own with them.

![](https://stomachache007.files.wordpress.com/2017/03/fullsizerender-12-03-17-23-26-2.jpeg?w=566\&h=600)

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();

        if (root == null) return res;

        // Divide
        List<Integer> left = new ArrayList<Integer>();
        List<Integer> right = new ArrayList<Integer>();

        // Conquer
        res.addAll(left);
        res.add(root.val);
        res.addAll(right);

        return res;
    }
}
```

## AC2: Traversal

Like a worker bee, traversal the tree and record in a note.

![](https://stomachache007.files.wordpress.com/2017/03/fullsizerender-12-03-17-23-26-3.jpeg?w=600\&h=344)

```java
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        traversal(root, res);
        return res;
    }

    private void traversal(TreeNode root, List<Integer> res) {
        if (root == null) return;

        traversal(root.left, res);
        res.add(root.val);
        traversal(root.right, res);
    }
}
```

## AC3: non-recursive

Use Stack to simulate recursion. In this way, it has higher memory capacity than system recursion.

```java
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // Variable
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode curr = root;

        // Corner case
        if (root == null) return res;

        // Business
        while (curr != null || !stack.empty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            res.add(curr.val);
            curr = curr.right;
        }

        return res;
    }

}
```