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[crumbs22] Week10 Solutions #1555

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Jun 8, 2025
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29 changes: 29 additions & 0 deletions invert-binary-tree/crumbs22.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,29 @@
/**
* 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) {
// 종료 조건
if (!root)
return (nullptr);

// 노드 반전 (중위순회)
TreeNode* tmp = root->left;
root->left = root->right;
root->right = tmp;

// 재귀적 호출
invertTree(root->left);
invertTree(root->right);
return (root);
}
};
35 changes: 35 additions & 0 deletions search-in-rotated-sorted-array/crumbs22.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,35 @@
class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size() - 1;
while(left <= right) {
int mid = left + (right - left) / 2;

if (nums[mid] == target)
return (mid);

// 왼쪽 구간이 정렬된 상태일 때
if (nums[left] <= nums[mid]) {
// 왼쪽 구간에 있으면 right를 좁히면서 target을 탐색
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
}
// 왼쪽 구간에 없다면 오른쪽 구간으로 이동
else
left = mid + 1;
}
// 오른쪽 구간이 정렬된 상태
else {
// 오른쪽 구간에 target이 있으면 left를 좁히면서 target 탐색
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
}
// 오른쪽 구간에 없다면 왼쪽 구간으로 이동
else
right = mid - 1;
}
}
return (-1);
}
};