leetcode2389--感染二叉树需要的总时间

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1. 题意

给定一个节点，每秒该节点会感染相邻的节点，受感染的节点下一秒也会感染周围节点；

求使得所有节点感染的时间

2. 题解

2.1 dfs建图+bfs搜索层次

我们将目标节点找到，并从该节点出发找到以该节点形成的树的深度即可。

我的代码

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {public:// * transform tree to graph// * sparse graphvoid build_graph(TreeNode * root, vector<vector<int>> &g) {if (nullptr == root)return;if ( root->left) {g[root->val].push_back( root->left->val);g[root->left->val].push_back(root->val);build_graph(root->left, g);}if ( root->right ) {g[root->val].push_back(root->right->val);g[root->right->val].push_back(root->val);build_graph(root->right, g);}}int amountOfTime(TreeNode* root, int start) {const int MAXN = 1e5 + 1;vector<vector<int>> g(MAXN, vector<int>());build_graph(root, g);int ans = 0;int last = start;queue<int> q;q.push(start);vector<int> vis(MAXN, 0);while (!q.empty()) {int cur = q.front();q.pop();vis[cur] = 1;for (int v:g[cur])if (!vis[v])q.push(v);if (cur == last) {ans++;last = q.back();}}return ans - 1;}
};

2.2 dfs建图+dfs求深度

我们需要找到每个节点的父节点，并且用一个from来表示当前节点是从哪来的以避免重复遍历。

我的代码

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {public:// * transform tree to graph// * sparse graphvoid get_fa(unordered_map<int,TreeNode *> &fa, TreeNode *root, int start, TreeNode *&start_node) {if (root == nullptr)return;if ( root->left)fa[root->left->val] = root;if ( root->right)fa[root->right->val] = root;if ( root->val == start)start_node = root;get_fa(fa, root->left, start, start_node );get_fa(fa, root->right, start, start_node );}int getDepth(TreeNode *root, int from,const unordered_map<int,TreeNode*> &fa) {if ( nullptr == root)return -1;int llen = -1;int rlen = -1;int flen = -1;if ( root->left && root->left->val != from)llen = getDepth(root->left, root->val, fa);if ( root->right && root->right->val != from )rlen = getDepth( root->right, root->val, fa);if (fa.find(root->val) != fa.end() &&fa.at(root->val)->val != from) {// cout << "fa-" << fa.at(root->val)->val << endl;flen = getDepth( fa.at(root->val), root->val, fa);}int ans = max(llen, rlen);ans = max(flen, ans);return ans + 1;}int amountOfTime(TreeNode* root, int start) {unordered_map<int,TreeNode *> fa;TreeNode *start_node;get_fa( fa, root, start, start_node);return  getDepth(start_node, -1, fa);}
};

03xf题解


TreeNode* fa[100001]; // 哈希表会超时，改用数组class Solution {int start;TreeNode* start_node;void dfs(TreeNode* node, TreeNode* from) {if (node == nullptr) {return;}fa[node->val] = from; // 记录每个节点的父节点if (node->val == start) {start_node = node; // 找到 start}dfs(node->left, node);dfs(node->right, node);}int maxDepth(TreeNode* node, TreeNode* from) {if (node == nullptr) {return -1; // 注意这里是 -1，因为 start 的深度为 0}int res = -1;if (node->left != from) {res = max(res, maxDepth(node->left, node));}if (node->right != from) {res = max(res, maxDepth(node->right, node));}if (fa[node->val] != from) {res = max(res, maxDepth(fa[node->val], node));}return res + 1;}public:int amountOfTime(TreeNode* root, int start) {this->start = start;dfs(root, nullptr);return maxDepth(start_node, start_node);}
};作者：灵茶山艾府
链接：https://leetcode.cn/problems/amount-of-time-for-binary-tree-to-be-infected/solutions/2753470/cong-liang-ci-bian-li-dao-yi-ci-bian-li-tmt0x/comments/2287846/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

2.3 一次遍历

03xf

class Solution {int ans = 0, start;pair<int, bool> dfs(TreeNode* node) {if (node == nullptr) {return {0, false};}auto [l_len, l_found] = dfs(node->left);auto [r_len, r_found] = dfs(node->right);if (node->val == start) {// 计算子树 start 的最大深度// 注意这里和方法一的区别，max 后面没有 +1，所以算出的也是最大深度ans = max(l_len, r_len);return {1, true}; // 找到了 start}if (l_found || r_found) {// 只有在左子树或右子树包含 start 时，才能更新答案ans = max(ans, l_len + r_len); // 两条链拼成直径// 保证 start 是直径端点return {(l_found ? l_len : r_len) + 1, true};}return {max(l_len, r_len) + 1, false};}public:int amountOfTime(TreeNode* root, int start) {this->start = start;dfs(root);return ans;}
};作者：灵茶山艾府
链接：https://leetcode.cn/problems/amount-of-time-for-binary-tree-to-be-infected/solutions/2753470/cong-liang-ci-bian-li-dao-yi-ci-bian-li-tmt0x/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。