1110. Delete Nodes And Return Forest

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1110. Delete Nodes And Return Forest

Description

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:

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

Example 2:

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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.

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * 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:
    set<int> s;
    vector<TreeNode*> res;

    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        s = set<int>(to_delete.begin(), to_delete.end());
        deleteNode(root, s, false);
        return res;
    }

    TreeNode* deleteNode(TreeNode* root, set<int>& s, bool hasParent) {
        if (!root) {
            return root;
        }
        bool deleted = s.contains(root->val);
        if (!deleted && !hasParent) {
            res.push_back(root);
        }
        root->left = deleteNode(root->left, s, !deleted);
        root->right = deleteNode(root->right, s, !deleted);
        return deleted ? nullptr : root;
    }
};