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37 changes: 37 additions & 0 deletions binary-tree-level-order-traversal/Jeehay28.js
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Queue 를 사용한 BFS 방식으로 문제를 접근해서 풀어주셨네요!
DFS를 사용한 문제 풀이 방법도 도전해보시면 재미있을것 같습니다 ㅎㅎ

Original file line number Diff line number Diff line change
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// ✅ Time Complexity: O(n), where n is the number of nodes in the tree
// ✅ Space Complexity: O(n)

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
if (!root) return [];

let output = [];
let queue = [root];

while (queue.length > 0) {
let values = [];

const lengthQueue = queue.length;

for (let i = 0; i < lengthQueue; i++) {
const node = queue.shift();
values.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
output.push(values);
}
return output;
};

51 changes: 51 additions & 0 deletions counting-bits/Jeehay28.js
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// ✅ Time Complexity: O(n)
// ✅ Space Complexity: O(n)

/**
* @param {number} n
* @return {number[]}
*/
var countBits = function (n) {
// dp[i] = 1 + dp[i - MSB]
// MSB(Most Significant Bit)

let dp = new Array(n + 1).fill(0);
let msb = 1;

for (let i = 1; i <= n; i++) {
if (msb * 2 === i) {
msb = i;
}

dp[i] = 1 + dp[i - msb];
}

return dp;
};

// ✅ Time Complexity: O(n * logn)
// ✅ Space Complexity: O(n)

/**
* @param {number} n
* @return {number[]}
*/
// var countBits = function (n) {
// let result = [0];
// const count = (num) => {
// let cnt = 0;
// while (num !== 0) {
// cnt += num % 2;
// num = Math.floor(num / 2);
// }
// return cnt;
// };

// for (let i = 1; i <= n; i++) {
// const temp = count(i);
// result.push(temp);
// }

// return result;
// };