BST to Greater Sum Tree
A medium BST problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- BST
- Sheets
- 1
- Core for
- 2 roles
- Platform
- LeetCode
The problem
Given the root of a binary search tree, convert it to a greater sum tree where each node's value is replaced by the sum of all values greater than or equal to that node's value in the original tree.
Example 1
- Input
- [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]
- Why
- Each node gets the sum of all values >= itself. For example, node 0 gets 0+1+2+3+4+5+6+7+8=36, node 4 gets 4+5+6+7+8=30.
Example 2
- Input
- [0,null,1]
- Output
- [1,null,1]
- Why
- Node 0 becomes 0+1=1, node 1 stays 1.
Example 3
- Input
- [1,0,2]
- Output
- [3,3,2]
- Why
- Node 0 becomes 0+1+2=3, node 1 becomes 1+2=3, node 2 stays 2.
Constraints
- The number of nodes is in the range [1, 104].
- 0 <= Node.val <= 100
- All values are unique.
How to think about it
Updated 2026-09-09Every node needs the sum of all values greater than or equal to itself. Standard inorder visits from smallest to largest, which is the wrong direction; reversing the visit order to right-root-left processes elements from largest to smallest. In that order, a single running sum accumulator carries everything each node needs.
Approaches, worst first
Inorder collect and suffix prefix array
time O(n) · space O(n)
Flatten the BST into a sorted array, compute prefix sums from the right, and perform a second tree pass to overwrite values from a hash table lookup. Correct, but wastes extra passes and auxiliary memory.
Reverse inorder recursion
time O(n) · space O(h)
Traverse right subtree, update running sum with current node value, overwrite current value with the running sum, and traverse left subtree. Visits each node once with stack depth bounded by tree height.
Reverse Morris traversalWrite this one
time O(n) · space O(1)
Establish temporary threads from each node's inorder successor back to the node to traverse in descending order without a stack. Updates the running accumulator in-place while achieving strictly constant extra memory.
Where people lose marks · 2
- Accumulating after overwriting. If you overwrite `node.val = sum` before adding the original value to `sum`, the running total double-counts or corrupts subsequent nodes.
- Traversing left before right. Standard inorder gives ascending order; updating with an accumulator during standard inorder computes smaller-or-equal sums instead of greater.
The theory behind it
BST — the ground this problem stands on. All BST problems
What BST is
A binary search tree is a binary tree that enforces a strict ordering rule at every node. Every value stored in a node's left branch must be smaller than the node itself, and every value in its right branch must be larger. Because this rule holds true everywhere down the tree, searching for a value does not require checking every branch. At each step, a single comparison lets you discard an entire half of the remaining nodes.
When to reach for it
Reach for a binary search tree when problems mention sorted tree structures, finding the kth smallest element, checking tree validity, or searching within a range of values. Signals include queries for the next greater element in a dynamic set, finding the lowest common ancestor in a sorted tree, or converting sorted arrays into height-balanced trees. When data must remain sorted while supporting continuous insertions and lookups, BST properties are the target.
How the pattern works
Use the ordering property to prune entire subtrees. When looking for a key, move left if the target is smaller than the current node, or move right if it is larger, stopping when values match or when reaching a null link. For validating a tree, do not merely compare a node to its immediate children; pass valid lower and upper value bounds down through recursive calls. Walking a valid binary search tree in-order visits all values in strictly increasing numerical order.
What each operation costs
| Operation | Time |
|---|---|
| search, insert, or delete on balanced tree | O(log n) |
| search, insert, or delete on degenerate line tree | O(n) |
| in-order traversal visiting all nodes in sorted order | O(n) |
What usually goes wrong with BST
- Validating a tree by checking only immediate children against the parent, missing violations where a node deep in the left subtree is larger than an ancestor higher up.
- Allowing duplicate values on the wrong side when the problem specification requires strictly smaller values on the left and strictly greater on the right.
- Deleting a node with two children by removing it directly instead of replacing its value with its in-order predecessor or successor and deleting that leaf instead.
Which roles need this problem
BST is a core topic for these 2 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 6 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
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