**Binary Search Tree Typo - Microsoft Top Interview Questions**

### 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);
values.add(node);
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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