From ceff0823b7d7bd661891d5ad1c14adf4c5e01ee6 Mon Sep 17 00:00:00 2001
From: Nishant Kaushal <101548649+nishant0708@users.noreply.github.com>
Date: Sat, 15 Jun 2024 20:28:22 +0530
Subject: [PATCH] Added Q538

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+---
+id: convert-bst-to-greater-tree
+title: Convert BST to Greater Tree
+sidebar_label: 0538-Convert-BST-to-Greater-Tree
+tags:
+- Tree
+- Depth-First Search
+- Binary Search Tree
+- Binary Tree
+description: "Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST."
+---
+
+## Problem
+
+Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]`  
+**Output:** `[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]`
+
+**Example 2:**
+
+**Input:** `root = [0,null,1]`  
+**Output:** `[1,null,1]`
+
+### Constraints
+
+- The number of nodes in the tree is in the range `[0, 10^4]`.
+- `-10^4 <= Node.val <= 10^4`
+- All the values in the tree are unique.
+- `root` is guaranteed to be a valid binary search tree.
+
+---
+
+## Approach
+
+To convert a BST to a Greater Tree, we need to accumulate the sum of all nodes that are greater than the current node. This can be efficiently done using a reverse in-order traversal (right-root-left), which processes nodes from the largest to the smallest.
+
+### Steps:
+
+1. Initialize a variable to keep track of the running sum.
+2. Perform a reverse in-order traversal of the tree.
+3. During the traversal, update the value of each node to include the running sum.
+4. Update the running sum with the value of the current node.
+
+### Solution
+
+#### Java  Solution
+
+```java
+public class TreeNode {
+    int val;
+    TreeNode left;
+    TreeNode right;
+    TreeNode(int x) { val = x; }
+}
+
+class Solution {
+    private int sum = 0;
+
+    public TreeNode convertBST(TreeNode root) {
+        if (root != null) {
+            convertBST(root.right);
+            sum += root.val;
+            root.val = sum;
+            convertBST(root.left);
+        }
+        return root;
+    }
+}
+```
+#### C++ Solution
+
+```cpp
+#include <iostream>
+using namespace std;
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+class Solution {
+private:
+    int sum = 0;
+public:
+    TreeNode* convertBST(TreeNode* root) {
+        if (root != nullptr) {
+            convertBST(root->right);
+            sum += root->val;
+            root->val = sum;
+            convertBST(root->left);
+        }
+        return root;
+    }
+};
+```
+#### Python Solution
+
+```python
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class Solution:
+    def convertBST(self, root: TreeNode) -> TreeNode:
+        self.sum = 0
+
+        def traverse(node):
+            if node:
+                traverse(node.right)
+                self.sum += node.val
+                node.val = self.sum
+                traverse(node.left)
+        
+        traverse(root)
+        return root
+```
+
+### Complexity Analysis
+**Time Complexity:** O(n)
+>Reason: Each node is visited once during the traversal.
+
+**Space Complexity:** O(h)
+>Reason: The space complexity is determined by the recursion stack, which in the worst case (unbalanced tree) is O(n), but on average (balanced tree) is O(log n).
+
+### References
+**LeetCode Problem:** Convert BST to Greater Tree
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