diff --git a/find-median-from-data-stream/forest000014.java b/find-median-from-data-stream/forest000014.java
new file mode 100644
index 000000000..e7f3f76c9
--- /dev/null
+++ b/find-median-from-data-stream/forest000014.java
@@ -0,0 +1,49 @@
+/*
+# Time Complexity: O(nlogn)
+# Space Complexity: O(nlogn)
+
+*/
+class MedianFinder {
+
+    private PriorityQueue<Integer> pq1; // 1번째 ~ 가운데 원소 (max heap)
+    private PriorityQueue<Integer> pq2; // 가운데+1번째 ~ 마지막 원소 (min heap)
+
+    public MedianFinder() {
+        pq1 = new PriorityQueue<>(Collections.reverseOrder());
+        pq2 = new PriorityQueue<>();
+    }
+
+    public void addNum(int num) {
+
+        if (pq1.size() == pq2.size()) {
+            if (pq1.peek() == null || pq1.peek() >= num) {
+                pq1.add(num);
+            } else {
+                pq2.add(num);
+                pq1.add(pq2.poll());
+            }
+        } else {
+            if (pq1.peek() == null || pq1.peek() >= num) {
+                pq1.add(num);
+                pq2.add(pq1.poll());
+            } else {
+                pq2.add(num);
+            }
+        }
+    }
+
+    public double findMedian() {
+        if (pq1.size() == pq2.size()) {
+            return (pq1.peek() + pq2.peek()) / 2.0;
+        } else {
+            return pq1.peek();
+        }
+    }
+}
+
+/**
+ * Your MedianFinder object will be instantiated and called as such:
+ * MedianFinder obj = new MedianFinder();
+ * obj.addNum(num);
+ * double param_2 = obj.findMedian();
+ */
diff --git a/insert-interval/forest000014.java b/insert-interval/forest000014.java
new file mode 100644
index 000000000..0488e75d3
--- /dev/null
+++ b/insert-interval/forest000014.java
@@ -0,0 +1,34 @@
+/*
+# Time Complexity: O(n)
+# Space Complexity: O(1)
+*/
+class Solution {
+    public int[][] insert(int[][] intervals, int[] newInterval) {
+        ArrayList<int[]> answer = new ArrayList<>();
+
+        int i = 0;
+        int n = intervals.length;
+
+        // newInterval보다 왼쪽의 구간들 추가
+        while (i < n && intervals[i][1] < newInterval[0]) {
+            answer.add(intervals[i]);
+            i++;
+        }
+
+        // newInterval과 겹치는 구간들 병합
+        while (i < n && intervals[i][0] <= newInterval[1]) {
+            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
+            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
+            i++;
+        }
+        answer.add(newInterval);
+
+        // newInterval보다 오른쪽의 구간들 추가
+        while (i < n) {
+            answer.add(intervals[i]);
+            i++;
+        }
+
+        return answer.toArray(new int[answer.size()][]);
+    }
+}
diff --git a/kth-smallest-element-in-a-bst/forest000014.java b/kth-smallest-element-in-a-bst/forest000014.java
new file mode 100644
index 000000000..c0ee7aea9
--- /dev/null
+++ b/kth-smallest-element-in-a-bst/forest000014.java
@@ -0,0 +1,53 @@
+/*
+# Time Complexity: O(n)
+# Space Complexity: O(n)
+
+heap을 사용한 풀이, in-order traversal을 사용한 풀이 풀어볼 것!
+*/
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public int kthSmallest(TreeNode root, int k) {
+        Map<TreeNode, Integer> sizes = new HashMap<>();
+
+        calculateSize(root, sizes);
+
+        return findKth(root, k, sizes);
+    }
+
+    private int calculateSize(TreeNode root, Map<TreeNode, Integer> sizes) {
+        if (root == null) return 0;
+
+        int left = calculateSize(root.left, sizes);
+        int right = calculateSize(root.right, sizes);
+
+        sizes.put(root, left + right + 1);
+        return left + right + 1;
+    }
+
+    private int findKth(TreeNode root, int k, Map<TreeNode, Integer> sizes) {
+        int left = (root.left == null) ? 0 : sizes.get(root.left);
+        int right = (root.right == null) ? 0 : sizes.get(root.right);
+
+        if (left == k - 1) {
+            return root.val;
+        } else if (left >= k) {
+            return findKth(root.left, k, sizes);
+        } else {
+            return findKth(root.right, k - left - 1, sizes);
+        }
+    }
+}
diff --git a/lowest-common-ancestor-of-a-binary-search-tree/forest000014.java b/lowest-common-ancestor-of-a-binary-search-tree/forest000014.java
new file mode 100644
index 000000000..b40d1544f
--- /dev/null
+++ b/lowest-common-ancestor-of-a-binary-search-tree/forest000014.java
@@ -0,0 +1,28 @@
+/*
+# Time Complexity: O(n)
+# Space Compleixty: O(1)
+*/
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        // null check unnecessary
+
+        if (root.val > p.val && root.val > q.val) {
+            return lowestCommonAncestor(root.left, p, q);
+        } else if (root.val < p.val && root.val < q.val) {
+            return lowestCommonAncestor(root.right, p, q);
+        } else {
+            return root;
+        }
+    }
+}
diff --git a/meeting-rooms/forest000014.java b/meeting-rooms/forest000014.java
new file mode 100644
index 000000000..56477d465
--- /dev/null
+++ b/meeting-rooms/forest000014.java
@@ -0,0 +1,15 @@
+/*
+# Time Complexity: O(nlogn)
+# Space Complexity: O(1)
+*/
+class Solution {
+    public boolean canAttendMeetings(int[][] intervals) {
+        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
+
+        for (int i = 1; i < intervals.length; i++) {
+            if (intervals[i][0] < intervals[i - 1][1]) return false;
+        }
+
+        return true;
+    }
+}