2026-06-01 18:12:40 +08:00
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#+ANKI_DECK: study_deck_02
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2026-06-01 17:12:10 +08:00
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* TODO 0226. Invert Binary Tree :easy:
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2026-06-01 02:33:30 +08:00
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:PROPERTIES:
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2026-06-01 17:22:07 +08:00
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:NEETCODE: [[file:../../roadmap.org::*0226. Invert Binary Tree][0226. Invert Binary Tree]]
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2026-06-01 02:33:30 +08:00
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:END:
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2026-06-01 17:22:07 +08:00
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Given the ~root~ of a binary tree, invert the tree, and return /its root/.
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*Example 1:*
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#+begin_src
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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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#+end_src
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*Example 2:*
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#+begin_src
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Input: root = [2,1,3]
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Output: [2,3,1]
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#+end_src
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*Example 3:*
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#+begin_src
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Input: root = []
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Output: []
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#+end_src
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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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2026-06-01 02:39:53 +08:00
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** TODO Approach
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Write your approach here.
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** TODO Python
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2026-06-05 22:32:49 +08:00
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#+begin_src python :lc-problem 226 :lc-lang python3
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2026-06-01 17:22:07 +08:00
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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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2026-06-01 02:39:53 +08:00
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#+end_src
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** TODO C++
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2026-06-05 22:32:49 +08:00
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#+begin_src cpp :lc-problem 226
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2026-06-01 17:22:07 +08:00
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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TreeNode* invertTree(TreeNode* root) {
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}
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};
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2026-06-01 02:33:30 +08:00
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#+end_src
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