1110. Delete Nodes And Return Forest

Problem Link

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:

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.

Solution

---

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:
    TreeNode* dfs(TreeNode* curr, bool isRoot, vector<TreeNode*>& res, unordered_set<int>& s) {
        if (!curr) return curr;
 
        bool shouldDelete = s.find(curr->val) != s.end();
 
        if (isRoot && !shouldDelete) {
            res.push_back(curr);
        }
 
        curr->left = dfs(curr->left, shouldDelete, res, s);
        curr->right = dfs(curr->right, shouldDelete, res, s);
 
        return shouldDelete ? NULL : curr;
    }
 
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        vector<TreeNode*> res;
        unordered_set<int> s(to_delete.begin(), to_delete.end());
 
        dfs(root, true, res, s);
 
        return res;
    }
};

Python3

# 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 delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:
        res = []
        delete = set(to_delete)
        
        def dfs(node, is_root):
            if not node: return None
            
            to_delete = node.val in delete
            
            if is_root and not to_delete:
                res.append(node)
            
            node.left = dfs(node.left, to_delete)
            node.right = dfs(node.right, to_delete)
            
            return None if to_delete else node
                        
        dfs(root, True)
        
        return res