226.Invert_Binary_Tree_(Easy).md

226. Invert Binary Tree (Easy)

Date and Time: May 31, 2024 (EST)
Update: Jul 15, 23:14 (EST)

Link: https://leetcode.com/problems/invert-binary-tree/

Question:

Given the root of a binary tree, invert the tree, and return its root.

Example 1:

Input: root = [4, 2, 7, 1, 3, 6, 9]

Output: [4, 7, 2, 9, 6, 3, 1]

Example 2:

Input: root = [2, 1, 3]

Output: [2, 3, 1]

Example 3:

Input: root = [ ]

Output: [ ]

Constraints:

• The number of nodes in the tree is in the range [0, 100].

• -100 <= Node.val <= 100

KeyPoints:

The way to solve this problem for both versions is: if root is not None (base case), we recursively set root.left, root.right = root.right, root.left.

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

Alternative 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root