LeetCode 1110. Delete Nodes And Return Forest

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.

实现思路：

将一棵树按照要求删除一个结点变成一片森林，模拟操作即可，注意细节。

AC代码：

class Solution {
		unordered_map<int,TreeNode*> mp;
		unordered_map<int,TreeNode*> father;//保留父节点
		bool vist[1010]= {false}; //是否是被删除的结点
	public:
		void dfs(TreeNode *root,TreeNode *fa) {
			if(!root) return;
			mp[root->val]=root;//保留结点
			dfs(root->left,root);
			dfs(root->right,root);
			father[root->val]=fa;//保留父节点
		}
		vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
			dfs(root,nullptr);
			for(int i=0; i<to_delete.size(); i++) {
				int val=to_delete[i];
				vist[val]=true;//当前结点已是被删除结点
				if(father[val]) {//当父节点不为空的时候才处理
					if(father[val]->left==mp[val]) father[val]->left=nullptr;//处理被删结点父节点
					else father[val]->right=nullptr;
				}
				if(mp[val]->left) father[mp[val]->left->val]=nullptr;//将当前删除结点和其孩子结点断开
				if(mp[val]->right) father[mp[val]->right->val]=nullptr;
			}
			vector<TreeNode*> ans;
			for(auto it : father) {
				if(vist[it.first]) continue;//如果是被删除的结点则不统计
				if(it.second==nullptr) ans.push_back(mp[it.first]);
			}
			return ans;
		}
};

方法二

后序遍历法

class Solution {//后序遍历法
	public:
		vector<TreeNode*> delNodes(TreeNode* root, vector<int>& toDelete) {
			vector<TreeNode*> result;
			unordered_set<int> worker;
			for (int &i : toDelete) worker.insert(i);
			delNodes(root, worker, result);
			if (root) result.push_back(root);
			return result;
		}

	private:
		void delNodes(TreeNode* &root, unordered_set<int>& toDelete, vector<TreeNode*> &result) {
			if (!root) return;
			delNodes(root->left, toDelete, result);
			delNodes(root->right, toDelete, result);
			if (toDelete.count(root->val)) {
				if (root->left) result.push_back(root->left);
				if (root->right) result.push_back(root->right);
				root = nullptr;
			}
		}
};