This problem requires converting a Binary Search Tree (BST) into a Greater Tree, where each node’s value is replaced by the sum of all greater values in the tree. A depth-first search approach with state tracking will be helpful to accumulate values as you traverse the tree from right to left.
Problem Statement
Given the root of a Binary Search Tree (BST), the task is to convert it into a Greater Tree. In the resulting tree, every node’s value will be updated to the sum of the original value plus the values of all nodes greater than it in the BST.
The conversion must be done efficiently, respecting the properties of the BST. A traversal method such as depth-first search (DFS) can help keep track of the cumulative sum of nodes processed so far while ensuring the tree structure is maintained.
Examples
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 details omitted.
Example 2
Input: root = [0,null,1]
Output: [1,null,1]
Example details omitted.
Constraints
- The number of nodes in the tree is in the range [0, 104].
- -104 <= Node.val <= 104
- All the values in the tree are unique.
- root is guaranteed to be a valid binary search tree.
Solution Approach
Depth-First Search (DFS) Traversal
A reverse in-order traversal (right to left) is used to ensure the sum of greater values is accumulated while visiting each node. This allows efficient updates of node values.
State Tracking for Accumulation
During the DFS traversal, maintain a variable that tracks the running total of values greater than the current node. Update each node’s value by adding this running total before moving to the left child.
In-place Tree Modification
The transformation happens in-place, meaning the tree structure is altered without the need for additional memory for a new tree. This ensures the solution is both space and time-efficient.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is O(N) where N is the number of nodes in the tree, as every node is visited once. The space complexity depends on the depth of recursion, O(H) where H is the height of the tree, which in the worst case is O(N).
What Interviewers Usually Probe
- Check if the candidate handles tree traversal correctly and efficiently.
- Assess how well the candidate manages state tracking during DFS traversal.
- Evaluate whether the candidate ensures the space complexity remains minimal by avoiding extra data structures.
Common Pitfalls or Variants
Common pitfalls
- Failing to handle the reverse in-order traversal properly, which leads to incorrect accumulation of greater values.
- Not updating the current node’s value before moving to its left child.
- Using unnecessary extra space or auxiliary data structures that increase space complexity unnecessarily.
Follow-up variants
- Implementing an iterative solution instead of recursion using a stack to simulate the DFS traversal.
- Optimizing the solution to handle very large trees with limited memory by using an iterative approach.
- Modifying the tree without recursion or stack, possibly by reordering the nodes in a non-recursive manner.
How GhostInterview Helps
- GhostInterview provides a structured approach to this problem, ensuring that you understand both the traversal method and state tracking needed for the solution.
- By solving this problem, you’ll gain deeper insights into efficient tree traversal and learn how to minimize space complexity in tree manipulations.
- GhostInterview’s hints guide you through best practices for maintaining and updating node values in-place while ensuring optimal time and space complexity.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
How does the in-order traversal help in solving the Convert BST to Greater Tree problem?
In-order traversal visits nodes in increasing order. For this problem, a reverse in-order (right to left) traversal is used to accumulate values of greater nodes first.
What is the time complexity of the Convert BST to Greater Tree problem?
The time complexity is O(N), where N is the number of nodes in the tree, as each node is visited once.
How do you avoid extra space when solving the Convert BST to Greater Tree problem?
By using in-place updates and performing the traversal recursively, you avoid the need for extra data structures, ensuring minimal space usage.
What is the pattern used in Convert BST to Greater Tree?
The problem uses binary-tree traversal with state tracking to accumulate the sum of greater nodes during the traversal.
Can the Convert BST to Greater Tree problem be solved iteratively?
Yes, an iterative solution can be implemented using a stack to simulate the DFS traversal, avoiding recursion.
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