/**
* 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<TreeNode *> delNodes(TreeNode *root, vector<int> &to_delete) {
unordered_set<int> deleteSet(to_delete.begin(), to_delete.end());
vector<TreeNode *> forest;
TreeNode *x = dfs(root, deleteSet, forest);
if (x != nullptr) {
forest.push_back(x);
}
return forest;
}
private:
TreeNode *dfs(TreeNode *root, unordered_set<int> &deleteSet, vector<TreeNode *> &forest) {
if (root == nullptr) {
return nullptr;
}
root->left = dfs(root->left, deleteSet, forest);
root->right = dfs(root->right, deleteSet, forest);
if (deleteSet.count(root->val) != 0) {
if (root->left != nullptr) {
forest.push_back(root->left);
}
if (root->right != nullptr) {
forest.push_back(root->right);
}
delete root;
return nullptr;
}
return root;
}
};
My Solution Solution 1: DFS (recursion) The Key Idea for Solving This Coding Question C++ Code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;
Jun 6, 2025MyCircularQueueO(k)
Apr 20, 2025O(m)
Mar 4, 2025O(n)n is the length of nums.
Feb 19, 2025or
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