[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 = [1]
Output: [1]

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