[leetcode ]429. N-ary Tree Level Order Traversale (easy)

it2025-11-18  11

原题

思路: bfs,每一层遍历一次加到一个vector,同时把该点的子元素加到queue中。

class Solution { public: vector<vector<int>> levelOrder(Node *root) { vector<vector<int>> res; if (root == NULL) return res; queue<Node *> q; q.push(root); while (!q.empty()) { vector<int> v; int len = q.size(); for (int i = 0; i < len; i++) { Node *temp = q.front(); q.pop(); v.push_back(temp->val); for (auto &a : temp->children) { q.push(a); } } res.push_back(v); } return res; } };

转载于:https://www.cnblogs.com/ruoh3kou/p/9893438.html

相关资源:数据结构—成绩单生成器
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