LeetCode 0559 Maximum Depth of N-ary Tree【N叉树，DFS】

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

---

题意

求出 N 叉树的深度

思路1

• DFS 保存一个全局最优解
• 根结点为空时需要特判

代码1

/*
// 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 {
    int max_depth = 0;
public:
    int maxDepth(Node* root) {
        if(root == NULL) return 0;
        dfs(root, 1);
        return max_depth;
    }
    
    void dfs(Node *s, int depth){
        if(s->children.size() == 0){
            max_depth = max(max_depth, depth);
            return;
        }
        for(auto &item : s->children)
            dfs(item, depth + 1);
    }
};