# LC 94. Binary Tree Inorder Traversal
### [Problem link](https://leetcode.com/problems/binary-tree-inorder-traversal/)
###### tags: `leedcode` `easy` `python` `c++` `Binary Tree`
Given the <code>root</code> of a binary tree, return the inorder traversal of its nodes' values.
**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg" style="width: 125px; height: 200px;" />
```
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 <code>[0, 100]</code>.
- <code>-100 <= Node.val <= 100</code>
**Follow up:** Recursive solution is trivial, could you do it iteratively?
## Solution 1
#### Python
##### recursive
```python=
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
def traversal(node: TreeNode):
if not node:
return
traversal(node.left)
res.append(node.val)
traversal(node.right)
traversal(root)
return res
```
##### iteration
```python=
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
stack = []
if root:
stack.append(root)
while stack:
node = stack.pop()
if node != None:
if node.right:
stack.append(node.right)
stack.append(node)
stack.append(None)
if node.left:
stack.append(node.left)
else:
node = stack.pop()
res.append(node.val)
return res
```
#### C++
##### recursive
```cpp=
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> res;
void traversal(TreeNode *cur) {
if (cur == nullptr) {
return;
}
traversal(cur->left);
res.push_back(cur->val);
traversal(cur->right);
}
vector<int> inorderTraversal(TreeNode* root) {
traversal(root);
return res;
}
};
```
##### iteration
```cpp=
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
if (root != nullptr) {
st.push(root);
}
while (!st.empty()) {
TreeNode *node = st.top();
st.pop();
if (node != nullptr) {
if (node->right) {
st.push(node->right);
}
st.push(node);
st.push(nullptr);
if (node->left) {
st.push(node->left);
}
} else {
node = st.top();
st.pop();
res.push_back(node->val);
}
}
return res;
}
};
```
>### Complexity
>n = number of tree's nodes.
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(n) |
## Note
[iteration reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E7%BB%9F%E4%B8%80%E8%BF%AD%E4%BB%A3%E6%B3%95.md)
[recursive reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E8%BF%AD%E4%BB%A3%E9%81%8D%E5%8E%86.md)