How to Solve Height of Binary Tree After Subtree Removal Queries Problem
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Height of Binary Tree After Subtree Removal Queries
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 ...
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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 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.
Height of Binary Tree After Subtree Removal Queries
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Examples
root = [1,3,4,2,null,6,5,null,null,null,null,null,7], queries = [4]
[2]
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).
root = [5,8,9,2,1,3,7,4,6], queries = [3,2,4,8]
[3,2,3,2]
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
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