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[Tony] WEEK 14 Solutions #592

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Nov 17, 2024
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26 changes: 26 additions & 0 deletions binary-tree-level-order-traversal/TonyKim9401.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,26 @@
// TC: O(n)
// need to visit all nodes
// SC: O(n)
// normally O(log n) required however, O(n) in the worst case
class Solution {
private List<List<Integer>> output = new ArrayList<>();
public List<List<Integer>> levelOrder(TreeNode root) {
dsf(0, root);
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(완전완전 사소한 typo) dsf -> dfs

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이런... 수정해두겠습니다 ㅠ 오타가 제일 무서워요

return output;
}

private void dsf(int level, TreeNode node) {
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문제를 보고 저는 자연스레 BFS가 떠올라서 BFS로 풀이했는데, 이렇게 DFS로 풀이할 수도 있군요 ㅎㅎ
잘 봤습니다 👍👍

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앗 저는 DFS밖에 생각이 안나서 ㅋㅋㅋ
Flynn님 풀이 보고 BFS 공부하겠습니다!

if (node == null) return;

if (output.size() == level) {
List<Integer> inside = new ArrayList<>();
inside.add(node.val);
output.add(inside);
} else {
output.get(level).add(node.val);
}
level += 1;
dsf(level, node.left);
dsf(level, node.right);
}
}
14 changes: 14 additions & 0 deletions reverse-bits/TonyKim9401.java
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
public class Solution {
public int reverseBits(int n) {
// time complexity O(1)
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엇 time complexity가 O(N) 아닌가요?

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for문의 반복 횟수가 Integer.SIZE = 32 로 고정되니 O(1) 이라고 생각했는데 아닐까요!?

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오 맞네요 제가 생각이 짧았습니다 ㅎㅎㅎ

// space complexity O(1)
int output = 0;

for (int i = 0; i < Integer.SIZE; i++) {
output <<= 1;
output += n & 1;
n >>= 1;
}
return output;
}
}