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1522. Diameter of N-Ary Tree

Description

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].

Solutions

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


    public Node() {
        children = new ArrayList<Node>();
    }

    public Node(int _val) {
        val = _val;
        children = new ArrayList<Node>();
    }

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

class Solution {
    private int ans;

    public int diameter(Node root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int dfs(Node root) {
        if (root == null) {
            return 0;
        }
        int m1 = 0, m2 = 0;
        for (Node child : root.children) {
            int t = dfs(child);
            if (t > m1) {
                m2 = m1;
                m1 = t;
            } else if (t > m2) {
                m2 = t;
            }
        }
        ans = Math.max(ans, m1 + m2);
        return 1 + m1;
    }
}

  • /*
// 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 ans;

    int diameter(Node* root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    int dfs(Node* root) {
        if (!root) return 0;
        int m1 = 0, m2 = 0;
        for (Node* child : root->children) {
            int t = dfs(child);
            if (t > m1) {
                m2 = m1;
                m1 = t;
            } else if (t > m2)
                m2 = t;
        }
        ans = max(ans, m1 + m2);
        return 1 + m1;
    }
};

  • """
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children if children is not None else []
"""


class Solution:
    def diameter(self, root: 'Node') -> int:
        """
        :type root: 'Node'
        :rtype: int
        """

        def dfs(root):
            if root is None:
                return 0
            nonlocal ans
            m1 = m2 = 0
            for child in root.children:
                t = dfs(child)
                if t > m1:
                    m2, m1 = m1, t
                elif t > m2:
                    m2 = t
            ans = max(ans, m1 + m2)
            return 1 + m1

        ans = 0
        dfs(root)
        return ans


  • /**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func diameter(root *Node) int {
	ans := 0
	var dfs func(root *Node) int
	dfs = func(root *Node) int {
		if root == nil {
			return 0
		}
		m1, m2 := 0, 0
		for _, child := range root.Children {
			t := dfs(child)
			if t > m1 {
				m2, m1 = m1, t
			} else if t > m2 {
				m2 = t
			}
		}
		ans = max(ans, m1+m2)
		return 1 + m1
	}
	dfs(root)
	return ans
}

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