LeetCode 226. Invert Binary Tree

파이썬 알고리즘 인터뷰 문제 45번(리트코드 226번) Invert Binary Tree
https://leetcode.com/problems/invert-binary-tree/

나의 풀이(재귀, recursive)

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

다른 풀이1(재귀, recursive)

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root:
            root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
            return root 

다른 풀이2(BFS, Queue)

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        queue = collections.deque([root])

        while queue:
            node = queue.popleft()
            if node:
                node.left, node.right = node.right, node.left
                queue.append(node.left)
                queue.append(node.right)
        return root

다른 풀이3(DFS, Stack, Pre-Order)

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        stack = [root]
        
        while stack:
            node = stack.pop()
            if node:
                node.left, node.right = node.right, node.left
                stack.append(node.left)
                stack.append(node.right)
        return root

다른 풀이4(DFS, Stack, Post-Order)

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        stack = [root]

        while stack:
            node = stack.pop()
            if node:
                stack.append(node.left)
                stack.append(node.right)
                node.left, node.right = node.right, node.left
        return root