559. Maximum Depth of N-ary Tree

Description

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.

Solutions

Solution 1: Recursion

First, we check if $\textit{root}$ is null. If it is, we return 0. Otherwise, we initialize a variable $\textit{mx}$ to record the maximum depth of the child nodes, then traverse all the child nodes of $\textit{root}$, recursively call the $\text{maxDepth}$ function, and update the value of $\textit{mx}$. Finally, we return $\textit{mx} + 1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes.

Python3JavaC++GoTypeScript

"""
# Definition for a Node.
class Node:
    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
        self.val = val
        self.children = children
"""


class Solution:
    def maxDepth(self, root: "Node") -> int:
        if root is None:
            return 0
        mx = 0
        for child in root.children:
            mx = max(mx, self.maxDepth(child))
        return 1 + mx

/*
// 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 mx = 0;
        for (Node child : root.children) {
            mx = Math.max(mx, maxDepth(child));
        }
        return 1 + mx;
    }
}

/*
// 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 mx = 0;
        for (Node* child : root->children) {
            mx = max(mx, maxDepth(child));
        }
        return mx + 1;
    }
};

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

func maxDepth(root *Node) int {
    if root == nil {
        return 0
    }
    mx := 0
    for _, child := range root.Children {
        mx = max(mx, maxDepth(child))
    }
    return 1 + mx
}

/**
 * Definition for _Node.
 * class _Node {
 *     val: number
 *     children: _Node[]
 *
 *     constructor(val?: number, children?: _Node[]) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = (children===undefined ? [] : children)
 *     }
 * }
 */

function maxDepth(root: _Node | null): number {
    if (!root) {
        return 0;
    }
    return 1 + Math.max(...root.children.map(child => maxDepth(child)), 0);
}

559. Maximum Depth of N-ary Tree

Description

Solutions

Solution 1: Recursion

Comments