Easy·[2025-12-23]
226. Invert Binary Tree
[#binary-tree, #bfs]
## description
## 226. Invert Binary Tree
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
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1Input: root = [4,2,7,1,3,6,9]
2Output: [4,7,2,9,6,3,1]Example 2:
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1Input: root = [2,1,3]
2Output: [2,3,1]Example 3:
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1Input: root = []
2Output: []- –The number of nodes in the tree is in the range [0, 100].
- –-100 <= Node.val <= 100
Constraints:
## notes
### Intuition
Inverting a tree means swapping every node's left and right children. We can traverse the tree using BFS (or DFS) and perform the swap at each node.
### Implementation
Handle the empty tree case first. Initialize a queue with the root. For each node dequeued, swap its left and right children. Then enqueue any non-null children to continue the traversal. Return the original root reference (now pointing to the inverted tree).
### Edge-cases
An empty tree (null root) should return null immediately. The algorithm handles single-node trees naturally.
- –Time: O(n) — visit each node once
- –Space: O(n) — queue holds up to n/2 nodes at the widest level
### Complexity
## solution
python
1from typing import Deque, Optional
2from lcutils import TreeNode
3from collections import deque
4
5# Definition for a binary tree node.
6# class TreeNode:
7# def __init__(self, val=0, left=None, right=None):
8# self.val = val
9# self.left = left
10# self.right = right
11class Solution:
12 def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
13 if root is None:
14 return root
15 q: Deque[TreeNode] = deque([root])
16
17 while q:
18 node = q.popleft()
19 node.right, node.left = node.right, node.left
20 if node.left:
21 q.append(node.left)
22 if node.right:
23 q.append(node.right)
24 return root
25
26
27
28if __name__ == "__main__":
29 # Include one-off tests here or debugging logic that can be run by running this file
30 # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
31 solution = Solution()
32--EOF--