559. 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].
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| /* | |
| // 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) { | |
| return dfs(root); | |
| } | |
| private int dfs(Node node) { | |
| if (node == null) { | |
| return 0; | |
| } | |
| int ans = 0; | |
| for (Node child: node.children) { | |
| ans = Math.max(ans, dfs(child)); | |
| } | |
| return 1 + ans; | |
| } | |
| } |
