LeetCode-1110. Delete Nodes And Return ForestLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/delete-nodes-and-return-forest/

Given the root of a binary tree, each node in the tree has a distinct value.

After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).

Return the roots of the trees in the remaining forest. You may return the result in any order.

Example 1:

Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
Output: [[1,2,null,4],[6],[7]]

Example 2:

Input: root = [1,2,4,null,3], to_delete = [3]
Output: [[1,2,4]]

Constraints:

• The number of nodes in the given tree is at most 1000.
• Each node has a distinct value between 1 and 1000.
• to_delete.length <= 1000
• to_delete contains distinct values between 1 and 1000.

 

【C++】

/**
 * 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 delNodes(TreeNode* root, vector<int>& to_delete) {
        vector forest;
        unordered_set<int> dict(to_delete.begin(), to_delete.end());
        root = helper(root, dict, forest);
        if (root) {
            forest.push_back(root);
        }
        return forest;
    }
    
    TreeNode* helper(TreeNode* root,
                    const unordered_set<int>& dict,
                    vector& forest) {
        if (!root) {return root;}
        root->left = helper(root->left, dict, forest);
        root->right = helper(root->right, dict, forest);
        if (dict.count(root->val)) {
            if (root->left) {forest.push_back(root->left);}
            if (root->right) {forest.push_back(root->right);}
            root = nullptr;
        }
        return root;
    }
};

【Java】

/**
 * 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 {
    public List delNodes(TreeNode root, int[] to_delete) {
        List forest = new ArrayList<>();
        Set dict = new HashSet() {{
            for (int num : to_delete) {add(num);}
        }};
        root = helper(root, dict, forest);
        if (root != null) {
            forest.add(root);
        }
        return forest;
    }
    
    TreeNode helper(TreeNode root,
                    Set dict,
                    List forest) {
        if (root == null) {return root;}
        root.left = helper(root.left, dict, forest);
        root.right = helper(root.right, dict, forest);
        if (dict.contains(root.val)) {
            if (root.left != null) {forest.add(root.left);}
            if (root.right != null) {forest.add(root.right);}
            root = null;
        }
        return root;
    }
}

参考文献

【1】Java中集合与数组之间的转换方法_跨过七海之风的博客-CSDN博客_java集合转数组的方法

【2】Java 几种方法初始化 Set_长安明月的博客-CSDN博客_java set初始化