✨ Day 31; Week 5 ✔️

Author: AkashRajpurohit

README.md

@@ -45,10 +45,11 @@
 6. [Linked List Cycle II](https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/562/week-4-october-22nd-october-28th/3509/) ➡️ [CPP Solution](Week4/detectCycle.cpp)
 7. [Summary Ranges](https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/562/week-4-october-22nd-october-28th/3510/) ➡️ [CPP Solution](Week4/summaryRanges.cpp)
 
-## Week 5 🚧
+## Week 5 ✅
 
 1. [Maximize Distance to Closest Person](https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/563/week-5-october-29th-october-31st/3512/) ➡️ [CPP Solution](Week5/maxDistToClosest.cpp)
 2. [Number of Longest Increasing Subsequence](https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/563/week-5-october-29th-october-31st/3513/) ➡️ [CPP Solution](Week5/findNumberOfLIS.cpp)
+3. [Recover Binary Search Tree](https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/563/week-5-october-29th-october-31st/3514/) ➡️ [CPP Solution](Week5/recoverTree.cpp)
 
 ## Other Challenges 💪
 

Week5/recoverTree.cpp

@@ -0,0 +1,41 @@
+/**
+ * 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* first = NULL;
+    TreeNode* second = NULL;
+    TreeNode* prev = new TreeNode(INT_MIN);
+    
+    void recoverTree(TreeNode* root) {
+        inorder(root);
+        
+        int temp = first->val;
+        first->val = second->val;
+        second->val = temp;
+    }
+    
+    void inorder(TreeNode* root) {
+        if(root == NULL) return;
+        
+        inorder(root->left);
+        
+        if(first == NULL && prev->val > root->val)
+            first = prev;
+        
+        if(first != NULL && prev->val > root->val)
+            second = root;
+        
+        prev = root;
+        
+        inorder(root->right);
+    }
+};