diff --git a/design-add-and-search-words-data-structure/lkhoony.js b/design-add-and-search-words-data-structure/lkhoony.js
new file mode 100644
index 0000000000..dfd61f78fb
--- /dev/null
+++ b/design-add-and-search-words-data-structure/lkhoony.js
@@ -0,0 +1,67 @@
+const Node = function() {
+    this.children = {};
+    this.isEnd = false;
+}
+
+var WordDictionary = function() {
+    this.root = new Node();
+};
+
+/**
+ * @param {string} word
+ * @return {void}
+ */
+WordDictionary.prototype.addWord = function(word) {
+    let currentNode = this.root;
+
+    for (let char of word) {
+        if (!currentNode.children[char]) {
+            currentNode.children[char] = new Node();
+        }
+        currentNode = currentNode.children[char];
+    }
+
+    currentNode.isEnd = true;
+};
+
+/**
+ * @param {string} word
+ * @return {boolean}
+ */
+WordDictionary.prototype.search = function(word) {
+    if (word === undefined) return false;
+    return this._search(this.root, word, 0);
+};
+
+WordDictionary.prototype._search = function(node, word, index) {
+    if (index === word.length) {
+        return node.isEnd;
+    }
+
+    const char = word[index];
+
+    if (char !== '.') {
+        const child = node.children[char];
+        return child ? this._search(child, word, index + 1) : false;
+    }
+
+    for (const key in node.children) {
+        if (this._search(node.children[key], word, index + 1)) {
+            return true;
+        }
+    }
+
+    return false;
+};
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * var obj = new WordDictionary()
+ * obj.addWord(word)
+ * var param_2 = obj.search(word)
+ */
+
+// n: 단어수, m: 단어 길이
+// addWord: 시간 복잡도 O(m)
+// search: 	시간 복잡도 O(m)
+// 공간 복잡도 O(n * m)
diff --git a/find-minimum-in-rotated-sorted-array/lkhoony.js b/find-minimum-in-rotated-sorted-array/lkhoony.js
new file mode 100644
index 0000000000..c980d6d6c1
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/lkhoony.js
@@ -0,0 +1,43 @@
+// math의 min을 이용하는 방법
+// tc: O(n^4)
+// sc: 잘 몰랐는데 모든 요소를 함수 인자로 풀어 콜스택에 올린다 하여 O(n)이 된다고 함..
+// (대용량의 배열 시 maximum exceed 에러가 날 수 있음)
+const findMin_use_math_min = function (nums) {
+  return Math.min(...nums);
+};
+
+// 메서드를 사용하지 않은 풀이.
+// tc: O(n^3)
+// sc: O(1)
+const findMin_naive = function (nums) {
+  let min = nums[0];
+
+  for (let i = 1; i < nums.length; i++) {
+    if (nums[i] <= min) {
+      min = nums[i];
+      break;
+    }
+  }
+
+  return min;
+};
+
+// 시간복잡도를 문제의 요구사항에 맞도록 줄여본 풀이
+// tc: O(n*logn)
+// sc: O(1)
+const findMin = function (nums) {
+  let left = 0,
+    right = nums.length - 1;
+
+  while (left < right) {
+    let mid = Math.floor((left + right) / 2);
+
+    if (nums[mid] > nums[right]) {
+      left = mid + 1;
+    } else {
+      right = mid;
+    }
+  }
+
+  return nums[left];
+};
diff --git a/graph-valid-tree/lkhoony.js b/graph-valid-tree/lkhoony.js
new file mode 100644
index 0000000000..961953e02b
--- /dev/null
+++ b/graph-valid-tree/lkhoony.js
@@ -0,0 +1,38 @@
+export class Solution {
+    /**
+     * @param {number} n - number of nodes
+     * @param {number[][]} edges - undirected edges
+     * @return {boolean}
+     */
+    validTree(n, edges) {
+      if (n === 0) return true;
+  
+      // 인접 리스트 생성
+      const adj = {};
+      for (let i = 0; i < n; i++) {
+        adj[i] = [];
+      }
+      for (const [n1, n2] of edges) {
+        adj[n1].push(n2);
+        adj[n2].push(n1);
+      }
+  
+      const visit = new Set();
+  
+      const dfs = (i, prev) => {
+        if (visit.has(i)) return false;
+  
+        visit.add(i);
+  
+        for (const j of adj[i]) {
+          if (j === prev) continue;
+          if (!dfs(j, i)) return false;
+        }
+  
+        return true;
+      };
+  
+      return dfs(0, -1) && visit.size === n;
+    }
+  }
+