C# Solution for Height of Binary Tree After Subtree Removal Queries Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to efficiently determine the maximum depth of a binary tree aft
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to efficiently determine the maximum depth of a binary tree after removing a node specified in the queries. Simply performing a depth-first search (DFS) for each query would be inefficient, especially for large trees and multiple queries. Instead, we can preprocess the tree to gather necessary information and then respond to each query in constant time.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tPreprocessing with DFS:\n\t\u2022\tWe perform a DFS on the tree to calculate two key pieces of information:\n\t\u2022\tNode Depths: Store the depth of each node in a list.\n\t\u2022\tNode Index Mapping: Create a mapping from each node\u2019s value to its index in the depth list.\n\t\u2022\tDuring the DFS, we also calculate the size of the subtree for each node, which tells us how many nodes are in the subtree rooted at that node.\n2.\tCalculating Maximum Depths:\n\t\u2022\tAfter the DFS, we construct two lists:\n\t\u2022\tMax Depth from Left: This maintains the maximum depth encountered when traversing from the root to the left of each node.\n\t\u2022\tMax Depth from Right: This is similar but maintains the maximum depth encountered when traversing from the root to the right.\n\t\u2022\tWe fill these lists by iterating through the node depths.\n3.\tProcessing Queries:\n\t\u2022\tFor each query, which specifies a node to be removed:\n\t\u2022\tDetermine the index of the node in the nodeDepths list.\n\t\u2022\tCalculate the maximum depth that can be achieved after removing that node by considering:\n\t\u2022\tThe maximum depth of the subtree on the left side of the node.\n\t\u2022\tThe maximum depth of the subtree on the right side of the node (if any).\n\t\u2022\tThis is done in constant time for each query, leveraging the preprocessed information.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tDFS Traversal: O(n) - Each node is visited once during the DFS to gather depths and subtree sizes.\n\u2022\tBuilding Maximum Depth Lists: O(n) - We compute the max depths for each node using the depth list.\n\u2022\tQuery Processing: O(1) for each query. If there are q queries, this step takes O(q).\n\nThus, the total time complexity is: O(n + q) \nwhere n is the number of nodes in the tree and q is the number of queries.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tNode Depths and Index Maps: O(n) - We use lists and dictionaries to store depths and subtree sizes, both of which scale with the number of nodes.\n\u2022\tAuxiliary Space: The DFS call stack may take O(h) space, where h is the height of the tree (in the worst case, this could be O(n) for skewed trees).\n\nThus, the total space complexity is: O(n) \n\n# Code\n```csharp []\n/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     public int val;\n *     public TreeNode left;\n *     public TreeNode right;\n *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\npublic class Solution {\n    public int[] TreeQueries(TreeNode root, int[] queries) {\n        // Maps to store the index of each node value and the number of nodes in subtree for each node\n        Dictionary<int, int> nodeIndexMap = new Dictionary<int, int>();\n        Dictionary<int, int> subtreeSize = new Dictionary<int, int>();\n\n        // Lists to store node depths and maximum depths from left and right\n        List<int> nodeDepths = new List<int>();\n        List<int> maxDepthFromLeft = new List<int>();\n        List<int> maxDepthFromRight = new List<int>();\n\n        // Perform DFS to populate nodeIndexMap and nodeDepths\n        DFS(root, 0, nodeIndexMap, nodeDepths);\n\n        int totalNodes = nodeDepths.Count;\n\n        // Calculate subtree sizes\n        CalculateSubtreeSize(root, subtreeSize);\n\n        // Calculate maximum depths from left and right\n        maxDepthFromLeft.Add(nodeDepths[0]);\n        maxDepthFromRight.Add(nodeDepths[totalNodes - 1]);\n\n        for (int i = 1; i < totalNodes; i++) {\n            maxDepthFromLeft.Add(Math.Max(maxDepthFromLeft[i - 1], nodeDepths[i]));\n            maxDepthFromRight.Add(Math.Max(maxDepthFromRight[i - 1], nodeDepths[totalNodes - i - 1]));\n        }\n        maxDepthFromRight.Reverse();\n\n        // Process queries\n        int[] results = new int[queries.Length];\n        for (int i = 0; i < queries.Length; i++) {\n            int queryNode = queries[i];\n            int startIndex = nodeIndexMap[queryNode] - 1;\n            int endIndex = startIndex + 1 + subtreeSize[queryNode];\n\n            int maxDepth = maxDepthFromLeft[startIndex];\n            if (endIndex < totalNodes) {\n                maxDepth = Math.Max(maxDepth, maxDepthFromRight[endIndex]);\n            }\n\n            results[i] = maxDepth;\n        }\n\n        return results;\n    }\n\n    // Depth-first search to populate nodeIndexMap and nodeDepths\n    private void DFS(TreeNode root, int depth, Dictionary<int, int> nodeIndexMap, List<int> nodeDepths) {\n        if (root == null) return;\n\n        nodeIndexMap[root.val] = nodeDepths.Count;\n        nodeDepths.Add(depth);\n\n        DFS(root.left, depth + 1, nodeIndexMap, nodeDepths);\n        DFS(root.right, depth + 1, nodeIndexMap, nodeDepths);\n    }\n\n    // Calculate the size of subtree for each node\n    private int CalculateSubtreeSize(TreeNode root, Dictionary<int, int> subtreeSize) {\n        if (root == null) return 0;\n\n        int leftSize = CalculateSubtreeSize(root.left, subtreeSize);\n        int rightSize = CalculateSubtreeSize(root.right, subtreeSize);\n\n        int totalSize = leftSize + rightSize + 1;\n        subtreeSize[root.val] = totalSize;\n\n        return totalSize;\n    }\n}\n```
['C#']