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+# Leetcode Invert-Binary-Tree
+
+#### 2022-07-06 13:33
+
+> ##### Algorithms:
+> #algorithm #DFS #recursion 
+> ##### Data structures:
+> #DS #binary_tree 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy
+> ##### Additional tags:
+> #leetcode
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#recursion tag:#DFS
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/invert-binary-tree/)
+___
+### Problem
+
+Given the `root` of a binary tree, invert the tree, and return _its root_.
+
+#### Examples
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/03/14/invert1-tree.jpg)
+
+```
+**Input:** root = [4,2,7,1,3,6,9]
+**Output:** [4,7,2,9,6,3,1]
+```
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/03/14/invert2-tree.jpg)
+
+```
+**Input:** root = [2,1,3]
+**Output:** [2,3,1]
+```
+
+**Example 3:**
+
+```
+**Input:** root = []
+**Output:** []
+```
+
+#### Constraints
+
+-   The number of nodes in the tree is in the range `[0, 100]`.
+-   `-100 <= Node.val <= 100`
+
+### Thoughts
+
+> [!summary]
+> This is a #DFS like recursion problem.
+
+Very simple, think of base cases:
+- the node is void, skip.
+And the flow is following
+
+- Catch base case
+- Invert sub-trees first
+- invert left and right node
+
+### Solution
+
+```cpp
+/**
+ * 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 {
+  void invert(TreeNode *root) {
+    if (!root) {
+      return;
+    }
+
+    invert(root->left);
+    invert(root->right);
+
+    TreeNode *tmp = root->left;
+    root->left = root->right;
+    root->right = tmp;
+  }
+
+public:
+  TreeNode *invertTree(TreeNode *root) {
+    // Using DFS-like Recursion
+    invert(root);
+    return root;
+  }
+};
+```
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