How do you solve Convert BST to Greater Tree on LeetCode?

Convert BST to Greater Tree (LeetCode #538) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The reverse in-order traversal allows us to efficiently calculate the cumulative sum of greater nodes.

What is the brute force solution for Convert BST to Greater Tree?

The brute force approach for Convert BST to Greater Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we traverse the tree for each node to find the sum of all greater nodes. This is straightforward but inefficient because we repeatedly traverse parts of the tree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Convert BST to Greater Tree?

Convert BST to Greater Tree uses the following concepts: Tree, Depth-First Search, Binary Search Tree, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Convert BST to Greater Tree?

Always clarify the problem requirements and constraints before diving into coding. Think about the time complexity of your approach and aim for the most efficient solution. Practice explaining your thought process as you code, as this is often part of the interview.

What are common mistakes when solving Convert BST to Greater Tree?

Not considering the order of traversal in a BST, which can lead to incorrect sums. Failing to maintain a running total during traversal, leading to repeated calculations.

#538

Convert BST to Greater Tree

Medium

TreeDepth-First SearchBinary Search TreeBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

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Intuition

Time O(n)Space O(h)

In the optimal approach, we perform a reverse in-order traversal of the BST. This way, we can keep a running total of the sum of all nodes we've visited so far, allowing us to update each node efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the cumulative sum (let's call it 'total').
2Step 2: Perform a reverse in-order traversal (right -> node -> left) of the tree.
3Step 3: For each node visited, update its value by adding the current 'total' to it, then update 'total' with the new value of the node.

solution.py19 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def convertBST(root):
9    total = 0
10    def reverse_inorder(node):
11        nonlocal total
12        if not node:
13            return
14        reverse_inorder(node.right)
15        total += node.val
16        node.val = total
17        reverse_inorder(node.left)
18    reverse_inorder(root)
19    return root

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1The reverse in-order traversal allows us to efficiently calculate the cumulative sum of greater nodes.
2Understanding the properties of a BST is crucial for optimizing tree-related problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.