Binary Tree Preorder Traversal

Improved Iterative 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        
        s = [root]
        res = []
        while s:
            node = s.pop()
            
            if node:
                res.append(node.val)
                s.append(node.right)
                s.append(node.left)
 
        return res

Iterative 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        
        s = [root]
        res = []
        while s != []:
            node = s[-1] #FIXME If popping, would it remove this element?
            # The answer is NO
            
            if node not in res:
                res.append(node)
 
            if node.left and (node.left not in res):
                s.append(node.left)
            else:
                popped = s.pop()
 
                # res.append(popped)
 
                if node.right and (node.right not in res):
                    s.append(node.right)
 
        takeOffPopped = []
        for node in res:
            takeOffPopped.append(node.val)
        
        return takeOffPopped

Recursive 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
 
        res = []
 
        def r(node):
            res.append(node.val)
 
            if node.left:
                r(node.left)
            
            if node.right:
                r(node.right)
 
            
        r(root)
        
        return res

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References

https://leetcode.com/problems/binary-tree-preorder-traversal/