题目
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.
For example, given a 3-ary tree:
假装是个图
We should return its max depth, which is 3.
Note:
The depth of the tree is at most 1000.
The total number of nodes is at most 5000.
图在这

代码
通过递归的思想,对子节点进行不同的处理
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val,List<Node> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public int maxDepth(Node root) {
if(root == null){
return 0;
}else{
return deep(root,1);
}
}
public static int deep(Node root ,int d){
List<Node> childer = root.children;
int size=childer==null?0:childer.size();
if(size==0){
return d;
}else{
int [] tag = new int[size];
for(int i=0;i<size;i++){
Node x = childer.get(i);
tag[i] = deep(x,d+1);
}
int max = 0;
for(int i=0;i<size;i++){
if(tag[i]>tag[max]){
max = i;
}
}
return tag[max];
}
}
}

本文介绍了一种递归算法来解决N叉树的最大深度问题。通过遍历树的每个子节点并递归地计算其深度,最终确定整棵树的最大深度。文章提供了完整的代码实现,并注明了树的深度不超过1000,节点总数不超过5000。
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