LeetCode 559. Maximum Depth of N-ary Tree

原题链接在这里：https://leetcode.com/problems/maximum-depth-of-n-ary-tree/description/

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.

题解：

Find the maxmimum depth among each subtree and return max + 1.

Time Complexity: O(n).

Space: O(h).

AC Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public int maxDepth(Node root) {
        if(root == null){
            return 0;
        }

        int depth = 0;
        for(Node next : root.children){
            depth = Math.max(depth, maxDepth(next));
        }

        return depth + 1;
    }
}

AC C++:

/*
// 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 depth = 0;
        for(Node* next : root->children){
            depth = max(depth, maxDepth(next));
        }

        return depth + 1;
    }
};