## February LeetCoding Challenge 2021

Day 9

## Convert BST to Greater Tree

**PROBLEM STATEMENT:**

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 sum of all keys greater than the original key in BST.
As a reminder, a *binary search tree*is a tree that satisfies these constraints:

- The left subtree of 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.

**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]

**Example 3**

Input:root = [1,0,2]Output:[3,3,2]

**Example 4**

Input:root = [3,2,4,1]Output:[7,9,4,10]

**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.

## Explanation

**ALGORITHM: **

- Given that the tree is a BST, we can find an O(n) solution. Traverse the tree in a reverse order. Reverse order traversal will encounter the nodes in decreasing order. Before visiting a node, we visit all greater nodes of that node and we will keep the track of sum of the node values which is the sum of all the keys greater than the key of current node.

## code

```
class Solution {
public:
void helper(TreeNode* root,int& val){
if(root->right)
helper(root->right,val);
root->val += val;
val = root->val;
if(root->left)
helper(root->left,val);
}
TreeNode* convertBST(TreeNode* root) {
if(!root) return root;
int val=0;
helper(root,val);
return root;
}
};
```