1038. Binary Search Tree to Greater Sum Tree

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.

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]

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• 0 <= Node.val <= 100
• All the values in the tree are unique.

Approach 01:

class Solution {
public:
    TreeNode* bstToGst(TreeNode* root) {
        int prefix_sum = 0;

        function<void(TreeNode*)> reversed_inorder_traversal = [&](TreeNode* node) {
                if (node == nullptr)
                    return;

                reversed_inorder_traversal(node->right);

                node->val += prefix_sum;
                prefix_sum = node->val;

                reversed_inorder_traversal(node->left);
            };

        reversed_inorder_traversal(root);
        return root;
    }
};

class Solution:
    def bstToGst(self, root: TreeNode) -> TreeNode:
        prefix_sum = 0

        def reversed_inorder_traversal(node: TreeNode):
            nonlocal prefix_sum
            if node is None:
                return

            reversed_inorder_traversal(node.right)

            node.val += prefix_sum
            prefix_sum = node.val

            reversed_inorder_traversal(node.left)

        reversed_inorder_traversal(root)
        return root

Time Complexity:

• The time complexity of the bstToGst function is \( O(n) \), where \( n \) is the number of nodes in the binary search tree.
• The reversed_inorder_traversal function visits each node exactly once, leading to a linear time complexity.

Space Complexity:

• The space complexity of the bstToGst function is \( O(h) \), where \( h \) is the height of the binary search tree.
• This space is used by the function call stack during the depth-first traversal (reversed inorder traversal).
• In the worst case, the space complexity can be \( O(n) \) if the tree is completely unbalanced (e.g., a skewed tree). For a balanced tree, the space complexity would be \( O(\log n) \).