# Invert a Binary Tree

Given the root of a binary tree, invert the tree, and return *its root*.

```
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root: #Base Case
return root
self.invertTree(root.left) #Call the left substree
self.invertTree(root.right) #Call the right substree
# Swap the nodes
root.left, root.right = root.right, root.left
return root # Return the root
```

## Related Problems

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the root of a binary tree, *determine if it is a valid binary search tree (BST)*.

A **valid BST** is defined as follows:

The left subtreeof a node contains only nodes with keys **less than** the node's key.

The right subtree of a node contains only nodes with keys **greater than** the node's key.

Both the left and right subtrees must also be binary search trees.

You are given the root of a binary search tree (BST), where the values of **exactly** two nodes of the tree were swapped by mistake. *Recover the tree without changing its structure*.