0094-Binary-Tree-Inorder-Traversal.py

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

Follow up: Recursive solution is trivial, could you do it iteratively?

Example 1:

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

Example 2:

Input: root = []
Output: []

Example 3:

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

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

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
'''

# 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

# Recursive Solution

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        self.recursive(root, res)
        return res
    
    def recursive(self, root, res):
        if not root:
            return
        self.recursive(root.left, res)
        res.append(root.val)
        self.recursive(root.right, res)

# Iterative Solution

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        if not root:
            return res
        stack = []
        node = root
        while node or stack:
            if node:
                stack.append(node)
                node = node.left
            else:
                new_node = stack.pop()
                res.append(new_node.val)
                node = new_node.right
            
        return res