题解 | #重量级的一层#
重量级的一层
https://www.nowcoder.com/practice/193372871b09426ab9ea805f0fd44d5c
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param root TreeNode类 * @return int整型 */ int maxLevelSum(TreeNode* root) { // write code here if (root == nullptr) return 0; queue<TreeNode*> q; q.push(root); int max_sum = 0, level = 0; int level_c = 0; while (q.size()) { level_c++; int size = q.size(); int sum = 0; for (int i = 0; i < size; i++) { TreeNode* t = q.front(); sum += t->val; if (t->left) q.push(t->left); if (t->right) q.push(t->right); q.pop(); } if (sum > max_sum) { level = level_c; max_sum = sum; } else if (sum == max_sum) { level = level_c; } } return level; } };