Formatted question description: https://leetcode.ca/all/1110.html

1110. Delete Nodes And Return Forest (Medium)

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.

Related Topics:
Tree, Depth-first Search

Solution 1.

// OJ: https://leetcode.com/problems/delete-nodes-and-return-forest/
// Time: O(N)
// Space: O(D + H)
class Solution {
    vector<TreeNode*> ans;
    unordered_set<int> s;
    void dfs(TreeNode *root, TreeNode *p = NULL) {
        if (!root) return;
        dfs(root->left, root);
        dfs(root->right, root);
        if (s.count(root->val)) {
            if (p != NULL) {
                if (p->left == root) p->left = NULL;
                else p->right = NULL;
            }
            if (root->left) ans.push_back(root->left);
            if (root->right) ans.push_back(root->right);
        } else if (p == NULL) ans.push_back(root);
    }
public:
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        s = unordered_set<int>(begin(to_delete), end(to_delete));
        dfs(root);
        return ans;
    }
};

/**
 * 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) {
        if (root == null)
            return new ArrayList<TreeNode>();
        Set<Integer> deleteSet = new HashSet<Integer>();
        for (int value : to_delete)
            deleteSet.add(value);
        List<TreeNode> nodesList = new ArrayList<TreeNode>();
        Map<TreeNode, TreeNode> map = new HashMap<TreeNode, TreeNode>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            nodesList.add(node);
            TreeNode left = node.left, right = node.right;
            if (left != null) {
                map.put(left, node);
                queue.offer(left);
            }
            if (right != null) {
                map.put(right, node);
                queue.offer(right);
            }
        }
        List<TreeNode> forest = new ArrayList<TreeNode>();
        if (!deleteSet.contains(root.val))
            forest.add(root);
        for (int i = nodesList.size() - 1; i >= 0 && !deleteSet.isEmpty(); i--) {
            TreeNode node = nodesList.get(i);
            if (deleteSet.contains(node.val)) {
                TreeNode parent = map.get(node);
                if (parent != null) {
                    if (node == parent.left)
                        parent.left = null;
                    else if (node == parent.right)
                        parent.right = null;
                }
                TreeNode left = node.left, right = node.right;
                if (left != null)
                    forest.add(left);
                if (right != null)
                    forest.add(right);
                deleteSet.remove(node.val);
            }
        }
        return forest;
    }
}

############

/**
 * 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) {
        boolean[] del = new boolean[1001];
        for (int d : to_delete) {
            del[d] = true;
        }
        List<TreeNode> res = new ArrayList<>();
        dfs(root, true, del, res);
        return res;
    }

    private TreeNode dfs(TreeNode root, boolean isRoot, boolean[] del, List<TreeNode> res) {
        if (root == null) {
            return null;
        }
        boolean flag = del[root.val];
        if (!flag && isRoot) {
            res.add(root);
        }
        root.left = dfs(root.left, flag, del, res);
        root.right = dfs(root.right, flag, del, res);
        return flag ? null : root;
    }
}

// OJ: https://leetcode.com/problems/delete-nodes-and-return-forest/
// Time: O(N)
// Space: O(D + H)
class Solution {
    vector<TreeNode*> ans;
    unordered_set<int> s;
    void dfs(TreeNode *root, TreeNode *p = NULL) {
        if (!root) return;
        dfs(root->left, root);
        dfs(root->right, root);
        if (s.count(root->val)) {
            if (p != NULL) {
                if (p->left == root) p->left = NULL;
                else p->right = NULL;
            }
            if (root->left) ans.push_back(root->left);
            if (root->right) ans.push_back(root->right);
        } else if (p == NULL) ans.push_back(root);
    }
public:
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        s = unordered_set<int>(begin(to_delete), end(to_delete));
        dfs(root);
        return ans;
    }
};

# 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]:
        def dfs(fa, root):
            if root is None:
                return
            dfs(root, root.left)
            dfs(root, root.right)
            if root.val in s:
                if fa and fa.left == root:
                    fa.left = None
                if fa and fa.right == root:
                    fa.right = None
                if root.left:
                    ans.append(root.left)
                if root.right:
                    ans.append(root.right)

        s = set(to_delete)
        ans = []
        if root.val not in s:
            ans.append(root)
        dfs(None, root)
        return ans

############

# 1110. Delete Nodes And Return Forest
# https://leetcode.com/problems/delete-nodes-and-return-forest/

# 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

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func delNodes(root *TreeNode, to_delete []int) []*TreeNode {
	s := map[int]bool{}
	for _, v := range to_delete {
		s[v] = true
	}
	ans := []*TreeNode{}
	if !s[root.Val] {
		ans = append(ans, root)
	}
	var fa *TreeNode
	var dfs func(fa, root *TreeNode)
	dfs = func(fa, root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root, root.Left)
		dfs(root, root.Right)
		if s[root.Val] {
			if fa != nil && fa.Left == root {
				fa.Left = nil
			}
			if fa != nil && fa.Right == root {
				fa.Right = nil
			}
			if root.Left != nil {
				ans = append(ans, root.Left)
			}
			if root.Right != nil {
				ans = append(ans, root.Right)
			}
		}
	}
	dfs(fa, root)
	return ans
}

1110 - Delete Nodes And Return Forest

1110. Delete Nodes And Return Forest (Medium)

Solution 1.

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1110. Delete Nodes And Return Forest (Medium)

Solution 1.

All Problems

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