559.Maximum Depth of N-ary Tree

Tags: Easy Tree

Links: https://leetcode.com/problems/maximum-depth-of-n-ary-tree/

---

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].

---

/*
// 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) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;
        auto &v = root -> children;
        int res = 0;
        for (auto e : v) {
            res = max(res, maxDepth(e));
        }

        return 1 + res;
    }
};

迭代解法：

/*
// 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) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;
        queue<Node *> q;
        q.push(root);
        int level = 0;
        while (!q.empty()) {
            int n = q.size();
            ++level;
            for (int i = 0; i < n; ++i) {
                Node *tmp = q.front(); q.pop();
                auto &v = tmp -> children;
                for (auto e : v) q.push(e);
            }
        }

        return level;
    }
};