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429. N-ary Tree Level Order Traversal**

429. N-ary Tree Level Order Traversal**

​​https://leetcode.com/problems/n-ary-tree-level-order-traversal/​​

题目描述

Given an n-ary tree, return the level order traversal of its nodes’ values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Example 1:


429. N-ary Tree Level Order Traversal**_leetcode

Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]

Example 2:


429. N-ary Tree Level Order Traversal**_算法_02

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

Constraints:

  • The height of the​​n-ary​​ tree is less than or equal to 1000
  • The total number of nodes is between​​[0, 10^4]​

C++ 实现 1

层序遍历使用队列来实现.

/*
// Definition for a Node.
class Node {
public:
int val;
vector children;

Node() {}

Node(int _val) {
val = _val;
}

Node(int _val, vector _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public:
vector<vector<int>> levelOrder(Node* root) {
if (!root) return {};
queue<Node*> q;
q.push(root);
vector<vector<int>> res;
while (!q.empty()) {
auto size = q.size();
vector<int> tmp;
while (size --) {
auto r = q.front();
q.pop();
tmp.push_back(r->val);
for (auto &c : r->children)
q.push(c);
}
res.push_back(tmp);
}
return res;
}
};

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