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:
// 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;
}
};