LeetCode #2458 — HARD

Height of Binary Tree After Subtree Removal Queries

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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. You are also given an array queries of size m.

You have to perform m independent queries on the tree where in the i^th query you do the following:

Remove the subtree rooted at the node with the value queries[i] from the tree. It is guaranteed that queries[i] will not be equal to the value of the root.

Return an array answer of size m where answer[i] is the height of the tree after performing the i^th query.

Note:

The queries are independent, so the tree returns to its initial state after each query.
The height of a tree is the number of edges in the longest simple path from the root to some node in the tree.

Example 1:

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).

Example 2:

Input: root = [5,8,9,2,1,3,7,4,6], queries = [3,2,4,8]
Output: [3,2,3,2]
Explanation: We have the following queries:
- Removing the subtree rooted at node with value 3. The height of the tree becomes 3 (The path 5 -> 8 -> 2 -> 4).
- Removing the subtree rooted at node with value 2. The height of the tree becomes 2 (The path 5 -> 8 -> 1).
- Removing the subtree rooted at node with value 4. The height of the tree becomes 3 (The path 5 -> 8 -> 2 -> 6).
- Removing the subtree rooted at node with value 8. The height of the tree becomes 2 (The path 5 -> 9 -> 3).

Constraints:

The number of nodes in the tree is n.
2 <= n <= 10⁵
1 <= Node.val <= n
All the values in the tree are unique.
m == queries.length
1 <= m <= min(n, 10⁴)
1 <= queries[i] <= n
queries[i] != root.val

Step 01

Brute Force Baseline

Problem summary: You are given the root of a binary tree with n nodes. Each node is assigned a unique value from 1 to n. You are also given an array queries of size m. You have to perform m independent queries on the tree where in the ith query you do the following: Remove the subtree rooted at the node with the value queries[i] from the tree. It is guaranteed that queries[i] will not be equal to the value of the root. Return an array answer of size m where answer[i] is the height of the tree after performing the ith query. Note: The queries are independent, so the tree returns to its initial state after each query. The height of a tree is the number of edges in the longest simple path from the root to some node in the tree.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Tree

Example 1

[1,3,4,2,null,6,5,null,null,null,null,null,7]
[4]

Example 2

[5,8,9,2,1,3,7,4,6]
[3,2,4,8]

Core Insight

What unlocks the optimal approach

Try pre-computing the answer for each node from 1 to n, and answer each query in O(1).
The answers can be precomputed in a single tree traversal after computing the height of each subtree.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<TreeNode, Integer> d = new HashMap<>();
    private int[] res;

    public int[] treeQueries(TreeNode root, int[] queries) {
        f(root);
        res = new int[d.size() + 1];
        d.put(null, 0);
        dfs(root, -1, 0);
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = res[queries[i]];
        }
        return ans;
    }

    private void dfs(TreeNode root, int depth, int rest) {
        if (root == null) {
            return;
        }
        ++depth;
        res[root.val] = rest;
        dfs(root.left, depth, Math.max(rest, depth + d.get(root.right)));
        dfs(root.right, depth, Math.max(rest, depth + d.get(root.left)));
    }

    private int f(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = f(root.left), r = f(root.right);
        d.put(root, 1 + Math.max(l, r));
        return d.get(root);
    }
}

// Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func treeQueries(root *TreeNode, queries []int) (ans []int) {
	d := map[*TreeNode]int{}
	var f func(*TreeNode) int
	f = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		l, r := f(root.Left), f(root.Right)
		d[root] = 1 + max(l, r)
		return d[root]
	}
	f(root)
	res := make([]int, len(d)+1)
	var dfs func(*TreeNode, int, int)
	dfs = func(root *TreeNode, depth, rest int) {
		if root == nil {
			return
		}
		depth++
		res[root.Val] = rest
		dfs(root.Left, depth, max(rest, depth+d[root.Right]))
		dfs(root.Right, depth, max(rest, depth+d[root.Left]))
	}
	dfs(root, -1, 0)
	for _, v := range queries {
		ans = append(ans, res[v])
	}
	return
}

# Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
# 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]

// Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
// /**
//  * Definition for a binary tree node.
//  * public class TreeNode {
//  *     int val;
//  *     TreeNode left;
//  *     TreeNode right;
//  *     TreeNode() {}
//  *     TreeNode(int val) { this.val = val; }
//  *     TreeNode(int val, TreeNode left, TreeNode right) {
//  *         this.val = val;
//  *         this.left = left;
//  *         this.right = right;
//  *     }
//  * }
//  */
// class Solution {
//     private Map<TreeNode, Integer> d = new HashMap<>();
//     private int[] res;
// 
//     public int[] treeQueries(TreeNode root, int[] queries) {
//         f(root);
//         res = new int[d.size() + 1];
//         d.put(null, 0);
//         dfs(root, -1, 0);
//         int m = queries.length;
//         int[] ans = new int[m];
//         for (int i = 0; i < m; ++i) {
//             ans[i] = res[queries[i]];
//         }
//         return ans;
//     }
// 
//     private void dfs(TreeNode root, int depth, int rest) {
//         if (root == null) {
//             return;
//         }
//         ++depth;
//         res[root.val] = rest;
//         dfs(root.left, depth, Math.max(rest, depth + d.get(root.right)));
//         dfs(root.right, depth, Math.max(rest, depth + d.get(root.left)));
//     }
// 
//     private int f(TreeNode root) {
//         if (root == null) {
//             return 0;
//         }
//         int l = f(root.left), r = f(root.right);
//         d.put(root, 1 + Math.max(l, r));
//         return d.get(root);
//     }
// }

// Accepted solution for LeetCode #2458: Height of Binary Tree After Subtree Removal Queries
function treeQueries(root: TreeNode | null, queries: number[]): number[] {
    const ans: number[] = [];
    const levels: Map<number, [number, number][]> = new Map();
    const valToLevel = new Map<number, number>();

    const dfs = (node: TreeNode | null, level = 0): number => {
        if (!node) return level - 1;

        const max = Math.max(dfs(node.left, level + 1), dfs(node.right, level + 1));

        if (!levels.has(level)) {
            levels.set(level, []);
        }
        levels.get(level)?.push([max, node.val]);
        valToLevel.set(node.val, level);

        return max;
    };

    dfs(root, 0);

    for (const [_, l] of levels) {
        l.sort(([a], [b]) => b - a);
    }

    for (const q of queries) {
        const level = valToLevel.get(q)!;
        const maxes = levels.get(level)!;

        if (maxes.length === 1) {
            ans.push(level - 1);
        } else {
            const [val0, max0, max1] = [maxes[0][1], maxes[0][0], maxes[1][0]];
            const max = val0 === q ? max1 : max0;
            ans.push(max);
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n+m)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.