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Given a root of an N-ary tree, you need to compute the length of the diameter of the tree.

The diameter of an N-ary tree is the length of the longest path between any two nodes in the tree. This path may or may not pass through the root.

(Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value.)

 

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: 3
Explanation: Diameter is shown in red color.

Example 2:

Input: root = [1,null,2,null,3,4,null,5,null,6]
Output: 4

Example 3:

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: 7

 

Constraints:

  • The depth of the n-ary tree is less than or equal to 1000.
  • The total number of nodes is between [1, 104].

Companies:
Facebook, Microsoft

Related Topics:
Tree, Depth-First Search

Similar Questions:

Solution 1. DFS

// OJ: https://leetcode.com/problems/diameter-of-n-ary-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int dfs(Node *root) {
        if (!root) return 0;
        int first = 0, second = 0;
        for (auto &c : root->children) {
            int d = dfs(c);
            if (d > first) {
                second = first;
                first = d;
            } else if (d > second) second = d;
        }
        ans = max(ans, first + second);
        return 1 + first;
    }
public:
    int diameter(Node* root) {
        dfs(root);
        return ans;
    }
};