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1110.delete-nodes-and-return-forest.java
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/*
* @lc app=leetcode id=1110 lang=java
*
* [1110] Delete Nodes And Return Forest
*
* https://leetcode.com/problems/delete-nodes-and-return-forest/description/
*
* algorithms
* Medium (65.72%)
* Likes: 810
* Dislikes: 32
* Total Accepted: 44.5K
* Total Submissions: 67.5K
* Testcase Example: '[1,2,3,4,5,6,7]\n[3,5]'
*
* 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]]
*
*
*
* 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.
*
*/
// @lc code=start
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
List<TreeNode> result = new ArrayList<>();
Set<Integer> delete = new HashSet<>();
for (int d : to_delete) delete.add(d);
TreeNode node = dfs(root, delete, result);
if (node != null) result.add(node);
return result;
}
private TreeNode dfs(TreeNode root, Set<Integer> delete, List<TreeNode> result) {
if (root == null) return null;
root.left = dfs(root.left, delete, result);
root.right = dfs(root.right, delete, result);
if (delete.contains(root.val)) {
if (root.left != null) result.add(root.left);
if (root.right != null) result.add(root.right);
return null;
} else {
return root;
}
}
}
// @lc code=end