diff --git a/invert-binary-tree/jaejeong1.java b/invert-binary-tree/jaejeong1.java
new file mode 100644
index 000000000..d72cf940f
--- /dev/null
+++ b/invert-binary-tree/jaejeong1.java
@@ -0,0 +1,44 @@
+/**
+ * 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 {
+  // 풀이: child node로 내려가면서 재귀로 계속 left와 right를 바꿔준다.
+  // TC: O(N)
+  // SC: O(N)
+  public TreeNode invertTree(TreeNode root) {
+    return invert(root);
+  }
+
+  private TreeNode invert(TreeNode node) {
+    if (node == null) {
+      return node;
+    }
+
+    node = swap(node);
+
+    invert(node.left);
+    invert(node.right);
+
+    return node;
+  }
+
+  private TreeNode swap(TreeNode node) {
+    var temp = node.left;
+    node.left = node.right;
+    node.right = temp;
+
+    return node;
+  }
+}
diff --git a/maximum-depth-of-binary-tree/jaejeong1.java b/maximum-depth-of-binary-tree/jaejeong1.java
new file mode 100644
index 000000000..b2c060a28
--- /dev/null
+++ b/maximum-depth-of-binary-tree/jaejeong1.java
@@ -0,0 +1,41 @@
+/**
+ * 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 {
+  // 풀이 : DFS 탐색을 통해 리프 노드까지 탐색하면서 깊이를 계산한다.
+  // TC: O(N), SC: O(N)
+  int answer = 0;
+
+  public int maxDepth(TreeNode root) {
+    dfs(root, 1);
+
+    return answer;
+  }
+
+  private void dfs(TreeNode node, int depth) {
+    if (node == null) {
+      return;
+    }
+
+    if (depth > answer) {
+      answer = depth;
+    }
+
+    depth++;
+
+    dfs(node.left, depth);
+    dfs(node.right, depth);
+  }
+}
diff --git a/reorder-list/jaejeong1.java b/reorder-list/jaejeong1.java
new file mode 100644
index 000000000..6c090b8db
--- /dev/null
+++ b/reorder-list/jaejeong1.java
@@ -0,0 +1,54 @@
+import java.util.Stack;
+
+// Definition for singly-linked list.
+class ListNode {
+
+  int val;
+  ListNode next;
+
+  ListNode() {
+  }
+
+  ListNode(int val) {
+    this.val = val;
+  }
+
+  ListNode(int val, ListNode next) {
+    this.val = val;
+    this.next = next;
+  }
+}
+
+class Solution {
+
+  public void reorderList(ListNode head) {
+    // 풀이: 역순으로 저장할 스택에 node들을 넣고, 기존 node 1개/스택 node 1개씩 이어 붙인다
+    // 스택은 LIFO로 저장되기 때문에, 문제에서 요구하는 순서대로 reorderList를 만들 수 있다
+    // TC: O(N)
+    // SC: O(N)
+    Stack<ListNode> stack = new Stack<>();
+
+    var curNode = head;
+    while(curNode != null) {
+      stack.push(curNode);
+      curNode = curNode.next;
+    }
+
+    curNode = head;
+    var halfSize = stack.size() / 2; // 한번에 2개씩 연결하기때문에 절반까지만 돌면 됨
+    for (int i=0; i<halfSize; i++) {
+      var top = stack.pop();
+
+      var nextNode = curNode.next;
+      curNode.next = top;
+      top.next = nextNode;
+
+      curNode = nextNode;
+    }
+
+    // 만약 노드의 개수가 홀수면, 하나의 노드가 남기 때문에 next 노드를 null로 처리해줘야 한다.
+    if (curNode != null) {
+      curNode.next = null;
+    }
+  }
+}