1110. Delete Nodes And Return Forest

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LeetCode

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Problem

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:

img

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.

Code

/**
 * 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<TreeNode> delNodes(TreeNode root, int[] to_delete) {
        Set<Integer> delete = new HashSet<>();
        for (int i : to_delete) delete.add(i);

        List<TreeNode> res = new ArrayList<>();
        if (!delete.contains(root.val)) res.add(root);

        dfs(root, delete, res);
        return res;
    }

    private TreeNode dfs(TreeNode node, Set<Integer> delete, List<TreeNode> res) {
        if (node == null) {
            return null;
        }

        node.left = dfs(node.left, delete, res);
        node.right = dfs(node.right, delete, res);

        if (delete.contains(node.val)) {
            if (node.left != null) res.add(node.left);
            if (node.right != null) res.add(node.right);
            return null;
        }

        return node;
    }
}