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sungjinwi.cpp
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50 lines (41 loc) · 1.28 KB
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/*
풀이 :
left와 right을 스왑하는 과정을 dfs를 이용해서 끝까지 반복한다
트리의 깊이 : N
TC : O(2^N)
깊이마다 2번씩 재귀함수 호출
SC : O(2^N)
깊이마다 2번씩 재귀함수 호출로 인한 스택 쌓임
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if (!root)
return nullptr;
TreeNode* tmp = root->left;
root->left = root->right;
root->right = tmp;
invertTree(root->left);
invertTree(root->right);
return root;
}
};
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};