[Leetcode 559]N叉树的最大深度Maximum Depth of N-ary Tree DFS/BFS模板

题目

https://leetcode.com/problems/maximum-depth-of-n-ary-tree/

N叉树的最大深度

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

思路

可以用BFS/DFS

BFS核心模板

Queue q=new LinkedList

q.add(初始点)

while(!q.isempty){

　　　q=q.size();

　　for(int i=0;i

　　　　cur=q.remove() 提出当前元素

　　　　广度优先遍历，不断加进新元素（q不为空时一直在while，为空时证明已经遍历完成）

　　　　q.add(初始点附近符合要求的节点)

　　这里写遍历过程中想记录更改什么

代码

BFS

    public int maxDepth(Node root) {
        //bfs
        if (root==null)
            return 0;
        int depth=0;

        Queue q=new LinkedList<>();
        q.offer(root);//加入初始点

        while(!q.isEmpty()){
            int size=q.size();
            for(int i=0;i){
                Node curNode=q.remove();//提取出当前节点
                for(Node child:curNode.children){
                    q.offer(child);//新加入满足条件的点
                }
            }
            depth++;
        }
        return depth;
    }

DFS

    int max_depth=0;
    public int maxDepth(Node root) {
        dfs(root,1);
        return max_depth;
    }
    
    public void dfs(Node node,int curDepth){
        if (node==null)
            return;
        
        max_depth=Math.max(max_depth,curDepth);
        
        for (Node child:node.children){
            dfs(child,curDepth+1);//每次depth+1
        }
        
    }