### Problem Statement#

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

#### Example 1:#

``````            4                           4
---->
2               7           7               2
1       3       6       9   9       6       3       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: []
``````

### Key Insights#

If you can swap every node’s left and right children, you can “invert” the binary tree. Once you see this, all you need to do is implement a simple post order binary tree traversal. The `recursive` and `iterative` approaches are shown below.

### Recursive Implementation#

``````# 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

from typing import Optional
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
# Handle null root case
if not root:
return root

if root.left:
self.invertTree(root.left)

if root.right:
self.invertTree(root.right)

# Swap left and right children
root.left, root.right = root.right, root.left

return root
``````

### Iterative Implementation#

``````# 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

from collections import deque
from typing import Optional
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
# Handle null root case
if not root:
return root

queue = deque([root])

while queue:
node = queue.popleft()

if node.left:
queue.append(node.left)

if node.right:
queue.append(node.right)

# Swap left and right children
node.left, node.right = node.right, node.left

return root
``````

Leetcode 226 Invert Binary Tree