题解 | #第k轻的牛牛#
第k轻的牛牛
https://www.nowcoder.com/practice/d3b31f055b1640d9b10de0a6f2b8e6f3
/** * 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 * @param k int整型 * @return int整型 */ int kthLighest(TreeNode* root, int k) { // write code here priority_queue<int, vector<int>, less<int>> q; queue<TreeNode*> que; que.push(root); while (que.size()) { int size = que.size(); for (int i = 0; i < size; i++) { TreeNode* t = que.front(); que.pop(); if (q.size() < k) { q.push(t->val); } else if (t->val < q.top()) { q.pop(); q.push(t->val); } if (t->left) que.push(t->left); if (t->right) que.push(t->right); } } return q.top(); } };