diff --git a/include/s0222_count_complete_tree_nodes.hpp b/include/s0222_count_complete_tree_nodes.hpp
new file mode 100644
index 0000000..20ccd8f
--- /dev/null
+++ b/include/s0222_count_complete_tree_nodes.hpp
@@ -0,0 +1,16 @@
+#ifndef S0222_COUNT_COMPLETE_TREE_NODES_HPP
+#define S0222_COUNT_COMPLETE_TREE_NODES_HPP
+
+#include <queue>
+
+#include "structures.hpp"
+
+using namespace std;
+
+class S0222 {
+ public:
+  int countNodes1(TreeNode* root);
+  int countNodes2(TreeNode* root);
+};
+
+#endif
diff --git a/src/s0222_count_complete_tree_nodes.cpp b/src/s0222_count_complete_tree_nodes.cpp
new file mode 100644
index 0000000..bbf7564
--- /dev/null
+++ b/src/s0222_count_complete_tree_nodes.cpp
@@ -0,0 +1,38 @@
+#include "s0222_count_complete_tree_nodes.hpp"
+
+// DFS O(n)
+int S0222::countNodes1(TreeNode* root) {
+  if (root == nullptr) return 0;
+  return countNodes1(root->left) + countNodes1(root->right) + 1;
+}
+
+// BFS O(n)
+int S0222::countNodes2(TreeNode* root) {
+  queue<TreeNode*> q;
+  int sum{0};
+  if (root) q.push(root);
+  TreeNode* front;
+  int size{0};
+  while (!q.empty()) {
+    size = q.size();
+    for (int i{0}; i < size; ++i) {
+      front = q.front();
+      q.pop();
+      ++sum;
+      if (front->left) q.push(front->left);
+      if (front->right) q.push(front->right);
+    }
+  }
+  return sum;
+}
+
+// 利用满二叉树的特性
+// 假设满二叉树一共有 h 层，那么总节点个数的范围应该是 [2^(h - 1), 2^h - 1] 之间
+// 也就是最后一层满和没满的区别
+// 所以可以在这个范围内用二分法搜索
+// 如何判断第 kkk 个节点是否存在呢？如果第 kkk 个节点位于第 hhh 层，则 kkk
+// 的二进制表示包含 h+1h+1h+1 位，其中最高位是
+// 111，其余各位从高到低表示从根节点到第 kkk 个节点的路径，000
+// 表示移动到左子节点，111 表示移动到右子节点。通过位运算得到第 kkk
+// 个节点对应的路径，判断该路径对应的节点是否存在，即可判断第 kkk
+// 个节点是否存在。