---
title: '226. Invert Binary Tree'
description: Given the root of a binary tree, invert the tree, and return its root
icon: dot
topics:
  - { name: "Tree", slug: "tree" }
  - { name: "Depth-First Search", slug: "depth-first-search" }
  - { name: "Breadth-First Search", slug: "breadth-first-search" }
  - { name: "Binary Tree", slug: "binary-tree" }
video: "OnSn2XEQ4MY"
issue: "https://github.com/prdlk/leetcode/issues/96"
companies:
  - { name: "Microsoft", period: "Thirty Days", frequency: 57.2 }
  - { name: "Google", period: "Thirty Days", frequency: 29 }
  - { name: "Meta", period: "Six Months", frequency: 32.7 }
sidebar:
  label: 'Invert Binary Tree'
  badge: 'Easy'
---

### 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`

## Approach

```mermaid
flowchart TD
  S(["invertTree(root)"]) --> B{"root is None?"}
  B -- yes --> Z(["return None"])
  B -- no --> L["left = invertTree(root.left), right = invertTree(root.right) — both subtrees are already inverted"]
  L --> X["root.left = right, root.right = left"]
  X --> E(["return root"])
```

## Solution

```py
# 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 None

        left = self.invertTree(root.left)
        right = self.invertTree(root.right)

        root.left = right
        root.right = left

        return root
```