###### tags: `leetcode` `easy` `Stack` `Tree` `Depth-First Search` `Binary Tree`
# [94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/description/)

## Description

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

## Examples
### Example1

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

### Example 2:

**Input**: root = []
**Output**: []

### Example 3:

**Input**: root = [1]
**Output**: [1]

## Constraints:

- The number of nodes in the tree is in the range `[0, 100]`.
- $-100 \leq Node.val \leq 100$

## Code

void inorder(int* returnList, struct TreeNode* root, int* returnSize) {
    if (root == NULL)
        return 0;
    inorder(returnList, root->left, returnSize);
    returnList[(*returnSize)++] = root->val;
    inorder(returnList, root->right, returnSize);
}

int* inorderTraversal(struct TreeNode* root, int* returnSize) {
    *returnSize = 0;
    int* returnList = malloc(100 * sizeof(int));
    inorder(returnList, root, returnSize);
    return returnList;
}

## Complexity

|Space |Time |
|- |- |
|$O(1)$|$O(N)$|

## Result

- Runtime : 0 ms, 100%
- Memory : 5.8 MB, 69.82%