Leetcode
2020.04.30 11:59

# 559. Maximum Depth of N-ary Tree

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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]`.

```/*
// 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) {
int depth = 0;

if(root == null){
return depth;
}

queue.offer(root);

while(queue.size() > 0){
int size = queue.size();
for(int i=0; i<size; i++){
Node node = queue.poll();
for(int j=0; j<node.children.size(); j++){
queue.offer(node.children.get(j));
}
}
depth++;
}

return depth;
}
}```

```/*
// 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) {
return findDepth(root, 1);
}

public int findDepth(Node node, int depth){
if(node == null){
return depth - 1;
}

if(node.children == null || node.children.size() == 0){
return depth;
}

int[] depths = new int[node.children.size()];
for(int i=0; i<depths.length; i++){
depths[i] = findDepth(node.children.get(i), depth + 1);
}

Arrays.sort(depths);

return depths[depths.length-1];
}
}```

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