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Formatted question description: https://leetcode.ca/all/2458.html

2458. Height of Binary Tree After Subtree Removal Queries

  • Difficulty: Hard.
  • Related Topics: Array, Tree, Depth-First Search, Breadth-First Search, Binary Tree.
  • Similar Questions: Maximum Depth of Binary Tree.

Problem

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.

  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

Solution (Java, C++, Python)

  • /**
 * 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);
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> treeQueries(TreeNode* root, vector<int>& queries) {
        unordered_map<TreeNode*, int> d;
        function<int(TreeNode*)> f = [&](TreeNode* root) -> int {
            if (!root) return 0;
            int l = f(root->left), r = f(root->right);
            d[root] = 1 + max(l, r);
            return d[root];
        };
        f(root);
        vector<int> res(d.size() + 1);
        function<void(TreeNode*, int, int)> dfs = [&](TreeNode* root, int depth, int rest) {
            if (!root) 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);
        vector<int> ans;
        for (int v : queries) ans.emplace_back(res[v]);
        return ans;
    }
};

  • # 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]


  • /**
 * 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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