Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.
Return the smallest level x such that the sum of all the values of nodes at level x is maximal.
Solution
直接用 \(queue\) 来 \(BFS\)， 遍历每一层更新答案
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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 {
private:
queue<TreeNode*> q;
int ans = INT_MIN;
int lev = 0;
public:
int maxLevelSum(TreeNode* root) {
q.push(root);
int real_lev=0;
while(!q.empty()){
int t = q.size();
int sum=0;
real_lev++;
while(t--){
auto f = q.front(); q.pop();
sum+= f->val;
if(f->left)q.push(f->left);
if(f->right)q.push(f->right);
}
//cout<<sum<<endl;
if(sum>ans){ans=sum;lev=real_lev;}
//cout<<lev<<endl;
}
return lev;
}
};