[LeetCode] 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].

多叉树的最大深度。这个题跟104题非常接近，唯一不同的是104题是二叉树，此题是多叉树。这个题可以用两种思路做，分别是BFS和DFS。两种做法的时间和空间复杂度都是O(n)。

BFS

JavaScript实现

/**
 * @param {Node} root
 * @return {number}
 */
var maxDepth = function (root) {
    if (root === null) return 0;
    let depth = 0;
    let queue = [root];
    while (queue.length) {
        let size = queue.length;
        for (let i = 0; i < size; i++) {
            let cur = queue.shift();
            queue.push(...cur.children);
        }
        depth++;
    }
    return depth;
};

Java实现

class Solution {
    public int maxDepth(Node root) {
        if (root == null) {
            return 0;
        }
        int depth = 0;
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Node cur = queue.poll();
                for (Node child : cur.children) {
                    queue.offer(child);
                }
            }
            depth++;
        }
        return depth;
    }
}

DFS

JavaScript实现

/**
 * @param {Node} root
 * @return {number}
 */
var maxDepth = function (root) {
    if (root === null) return 0;
    let res = 0;
    for (let child of root.children) {
        res = Math.max(res, maxDepth(child));
    }
    return res + 1;
};

Java实现

class Solution {
    public int maxDepth(Node root) {
        if (root == null) {
            return 0;
        }
        int res = 0;
        for (Node child : root.children) {
            res = Math.max(res, maxDepth(child));
        }
        return res + 1;
    }
}

LeetCode 题目总结