Leetcode #1038: Binary Search Tree to Greater Sum Tree
In this guide, we solve Leetcode #1038 Binary Search Tree to Greater Sum Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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 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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Binary Search Tree, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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]
Python Solution
# 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 bstToGst(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def dfs(root: Optional[TreeNode]):
if root is None:
return
dfs(root.right)
nonlocal s
s += root.val
root.val = s
dfs(root.left)
s = 0
dfs(root)
return root
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.