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文章目录
- 一、题目
- 二、题解
一、题目
Given a n-ary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
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:
Input: root = [1,null,3,2,4,null,5,6]
Output: 3
Example 2:
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: 5
Constraints:
The total number of nodes is in the range [0, 104].
The depth of the n-ary tree is less than or equal to 1000.
二、题解
/*
// Definition for a Node.
class Node {
public:int val;vector<Node*> children;Node() {}Node(int _val) {val = _val;}Node(int _val, vector<Node*> _children) {val = _val;children = _children;}
};
*/class Solution {
public:int maxDepth(Node* root) {if(!root) return 0;int res = 0;queue<Node*> q;q.push(root);while(!q.empty()){int size = q.size();while(size--){Node* t = q.front();q.pop();for(int i = 0;i < t->children.size();i++){q.push(t->children[i]);}}res++;}return res;}
};
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