# [Solved] Given the root of a binary tree, return the inorder traversal of its nodes’ values. ## Question

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

Example 1:

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

Example 2:

```Input: root = []
Output: []
```

Example 3:

```Input: root = 
Output: 
```

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?

## Python Solution

```# 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: TreeNode) -> List[int]:
stack = []
l = []
curr = root
while 1:
if curr is not None:
stack.append(curr)
curr = curr.left
elif stack:
x = stack.pop()
l.append(x.val)
curr = x.right
else:
break

return l```
Scroll to Top