#226 solution

Author: sungjinwi

invert-binary-tree/sunginwi.cpp

@@ -0,0 +1,50 @@
+/*
+    풀이 : 
+        left와 right을 스왑하는 과정을 dfs를 이용해서 끝까지 반복한다
+
+    트리의 깊이 : N
+
+    TC : O(2^N)
+        깊이마다 2번씩 재귀함수 호출
+
+    SC : O(2^N)
+        깊이마다 2번씩 재귀함수 호출로 인한 스택 쌓임
+*/
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+    public:
+        TreeNode* invertTree(TreeNode* root) {
+            if (!root)
+                return nullptr;
+    
+            TreeNode* tmp = root->left;
+    
+            root->left = root->right;
+            root->right = tmp;
+
+            invertTree(root->left);
+            invertTree(root->right);
+
+            return root;
+        }
+    };
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode() : val(0), left(nullptr), right(nullptr) {}
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+};