2385. 感染二叉树需要的总时间

2385. 感染二叉树需要的总时间

题目链接：2385. 感染二叉树需要的总时间

代码如下：

/**
 * 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) {}
 * };
 */
 //记录父节点 + DFS
 //参考链接：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
class Solution {
public:
    void dfs(TreeNode* node,TreeNode* from)
    {
        if(node==nullptr)   return;
        fa[node->val]=from;//记录每个结点的父节点
        if(node->val==start)//找到start节点
        {
            start_node=node;
        }
        dfs(node->left,node);
        dfs(node->right,node);
    }

    int maxDepth(TreeNode* node,TreeNode* from)
    {
        if(node==nullptr)   return -1;
        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;
    }

    int amountOfTime(TreeNode* root, int start) 
    {
        this->start=start;
        dfs(root,nullptr);
        return maxDepth(start_node,start_node);
    }

private:
    int start;
    TreeNode* start_node;
    TreeNode* fa[100001];
};