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2385. 感染二叉树需要的总时间.cpp
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2385. 感染二叉树需要的总时间.cpp
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/**
* 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:
void solve2(TreeNode* root, unordered_map<TreeNode*,TreeNode*>&childToParent,TreeNode* parent){
if(root==NULL){
return;
}
queue<TreeNode*>q;
q.push(root);
childToParent[root]=NULL;
while(!q.empty()){
TreeNode* front=q.front();
q.pop();
if(front->left){
childToParent[front->left]=front;
q.push(front->left);
}
if(front->right){
childToParent[front->right]=front;
q.push(front->right);
}
}
}
TreeNode* solve(TreeNode* root, int start){
if(root==NULL) return NULL;
if(root->val==start) return root;
TreeNode* left=solve(root->left,start);
TreeNode* right=solve(root->right,start);
if(left==right) return left;
if(left==NULL && right!=NULL) return right;
return left;
}
int amountOfTime(TreeNode* root, int start) {
queue<TreeNode*>q;
TreeNode* find=solve(root,start);
//cout<<find->val<<endl;
q.push(find);
unordered_map<int,bool>mapping;
unordered_map<TreeNode*,TreeNode*>childToParent;
solve2(root,childToParent,NULL);
// for(auto i : childToParent){
// cout<<i.first->val<<" ";
// if(i.second!=NULL){
// cout<<i.second->val<<endl;
// }else{
// cout<<"Null"<<endl;
// }
// }
int time=0;
while(!q.empty()){
int size=q.size();
for(int i=0;i<size;i++){
TreeNode* front=q.front();
q.pop();
mapping[front->val]=true;
if(front->left!=NULL && mapping.find(front->left->val)==mapping.end()){
q.push(front->left);
}
if(front->right!=NULL && mapping.find(front->right->val)==mapping.end()){
q.push(front->right);
}
TreeNode* parent=childToParent[front];
if(parent!=NULL && mapping.find(parent->val)==mapping.end()){
q.push(parent);
}
}
time++;
}
return time-1;
}
};