Leetcode #2458: Height of Binary Tree After Subtree Removal Queries
In this guide, we solve Leetcode #2458 Height of Binary Tree After Subtree Removal Queries 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
You are given the root of a binary tree with n nodes. Each node is assigned a unique value from 1 to n.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Array, 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 = [1,3,4,2,null,6,5,null,null,null,null,null,7], queries = [4]
Output: [2]
Explanation: The diagram above shows the tree after removing the subtree rooted at node with value 4.
The height of the tree is 2 (The path 1 -> 3 -> 2).
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 treeQueries(self, root: Optional[TreeNode], queries: List[int]) -> List[int]:
def f(root):
if root is None:
return 0
l, r = f(root.left), f(root.right)
d[root] = 1 + max(l, r)
return d[root]
def dfs(root, depth, rest):
if root is None:
return
depth += 1
res[root.val] = rest
dfs(root.left, depth, max(rest, depth + d[root.right]))
dfs(root.right, depth, max(rest, depth + d[root.left]))
d = defaultdict(int)
f(root)
res = [0] * (len(d) + 1)
dfs(root, -1, 0)
return [res[v] for v in queries]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.