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67 lines
1.6 KiB
Plaintext
67 lines
1.6 KiB
Plaintext
---
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title: '226. Invert Binary Tree'
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description: Given the root of a binary tree, invert the tree, and return its root
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icon: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Binary Tree", slug: "binary-tree" }
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video: "OnSn2XEQ4MY"
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issue: "https://github.com/prdlk/leetcode/issues/96"
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sidebar:
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label: 'Invert Binary Tree'
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badge: 'Easy'
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---
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### Example 1:
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- Input: `root = [4,2,7,1,3,6,9]`
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- Output: `[4,7,2,9,6,3,1]`
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### Example 2:
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- Input: `root = [2,1,3]`
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- Output: `[2,3,1]`
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### Example 3:
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- Input: `root = []`
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- Output: `[]`
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### Constraints:
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- The number of nodes in the tree is in the range [0, 100].
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- `-100 <= Node.val <= 100`
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## Approach
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```mermaid
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flowchart TD
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S(["invertTree(root)"]) --> B{"root is None?"}
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B -- yes --> Z(["return None"])
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B -- no --> L["left = invertTree(root.left), right = invertTree(root.right) — both subtrees are already inverted"]
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L --> X["root.left = right, root.right = left"]
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X --> E(["return root"])
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```
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## Solution
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```py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
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if not root:
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return None
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left = self.invertTree(root.left)
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right = self.invertTree(root.right)
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root.left = right
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root.right = left
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return root
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```
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