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.
思路：每次從result list裏遍歷所有root節點，找到要刪除的node及其父節點，把父節點的子節點node設爲null，添加node節點的左右孩子（如果不爲null的話）進result list。
class Solution {
public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
List<TreeNode> res = new ArrayList<>();
res.add(root);
for (int i = 0; i < to_delete.length; i++) {
TreeNode[] cur = new TreeNode[2];
for (int j = 0; j < res.size(); j++) {
cur = helper(res.get(j), to_delete[i]);
if (cur[1] != null) {
if (cur[0] == null) {
res.remove(cur[1]);
if (cur[1].left != null) res.add(cur[1].left);
if (cur[1].right != null) res.add(cur[1].right);
} else {
if (cur[0].left != null && cur[0].left.val == cur[1].val) {
cur[0].left = null;
} else {
cur[0].right = null;
}
if (cur[1].left != null) res.add(cur[1].left);
if (cur[1].right != null) res.add(cur[1].right);
}
}
}
}
return res;
}
public TreeNode[] helper(TreeNode root, int num) {
TreeNode[] res = new TreeNode[2];
if (root.val == num) {
res[1] = root;
return res;
} else if (root.left != null && root.left.val == num) {
res[0] = root;
res[1] = root.left;
} else if (root.right != null && root.right.val == num) {
res[0] = root;
res[1] = root.right;
} else {
if (root.left != null) res = helper(root.left, num);
if (res[1] != null) return res;
if (root.right != null) res = helper(root.right, num);
}
return res;
}
}