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 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.

Note: This question is the same as 538: https://leetcode.com/problems/convert-bst-to-greater-tree/


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]



  • 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.
  • root is guaranteed to be a valid binary search tree.

1038. Binary Search Tree to Greater Sum Tree
struct Solution;
use rustgym_util::*;

trait Inorder {
    fn inorder(&mut self, sum: &mut i32);

impl Inorder for TreeLink {
    fn inorder(&mut self, sum: &mut i32) {
        if let Some(node) = self {
            *sum += node.borrow().val;
            node.borrow_mut().val = *sum;

impl Solution {
    fn bst_to_gst(mut root: TreeLink) -> TreeLink {
        let mut sum = 0;
        root.inorder(&mut sum);

fn test() {
    let root = tree!(
        tree!(1, tree!(0), tree!(2, None, tree!(3))),
        tree!(6, tree!(5), tree!(7, None, tree!(8)))
    let res = tree!(
        tree!(36, tree!(36), tree!(35, None, tree!(33))),
        tree!(21, tree!(26), tree!(15, None, tree!(8)))
    assert_eq!(Solution::bst_to_gst(root), res);