### Problem Statement :

```You are given a binary tree root which is almost a binary search tree except two nodes' values have been swapped. Return the original binary search tree.

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [2, [5, null, null], [7, [1, null, null], [8, null, null]]]

Output

[2, [1, null, null], [7, [5, null, null], [8, null, null]]]

Explanation

We can swap 1 and 5.

Example 2

Input

root = [0, [1, null, null], null]

Output

[1, [0, null, null], null]

Explanation

We can swap 0 and 1.```

### Solution :

```                        ```Solution in C++ :

void check(Tree* root, Tree*& prev, Tree*& node1, Tree*& node2) {
if (!root) return;
check(root->left, prev, node1, node2);
if (prev && prev->val >= root->val) {
if (node1 == nullptr) node1 = prev;
node2 = root;
}
prev = root;
check(root->right, prev, node1, node2);
}

Tree* solve(Tree* root) {
Tree *node1 = nullptr, *node2 = nullptr, *prev = nullptr;
check(root, prev, node1, node2);
swap(node1->val, node2->val);
return root;
}```
```

```                        ```Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
public Tree solve(Tree root) {
List<Tree> values = new ArrayList<>();
inOrder(root, values);
findAndSwap(values);
return root;
}

private void inOrder(Tree node, List<Tree> values) {
if (node == null)
return;

inOrder(node.left, values);
inOrder(node.right, values);
}

private void findAndSwap(List<Tree> values) {
Tree first = null;
Tree second = null;

for (int i = 1; i < values.size(); i++) {
if (values.get(i - 1).val > values.get(i).val) {
if (first == null) {
first = values.get(i - 1);
second = values.get(i);
} else {
second = values.get(i);
}
}
}

int temp = first.val;
first.val = second.val;
second.val = temp;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
left = right = None
last_node = Tree(-float("inf"))

for node in self.morris_inorder(root):
if node.val < last_node.val:
left = left or last_node
right = node
last_node = node

left.val, right.val = right.val, left.val

return root

@staticmethod
def morris_inorder(node):
temp = None
while node:
if node.left:
temp = node.left
while temp.right and temp.right != node:
temp = temp.right
if temp.right:
temp.right = None
yield node
node = node.right
else:
temp.right = node
node = node.left
else:
yield node
node = node.right```
```

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