可以尝试使用递归和非递归两种方式解决

C++ 递归

class Solution {
public:
  int maxDepth(Node *root) {
    if (NULL == root) {
      return 0;
    }
    if (root->children.size() == 0) {
      return 1; //提升4ms，不需要做for循环判断
    }

    int depth = 0;
    for (int i = 0; i < root->children.size(); i++) {
      depth =
          max(depth,
              maxDepth(root->children[i])); //选择最大的一个，这里没有+1这动作
    }

    return depth + 1;
  }
};

c++ 非递归

class Solution_DFS {
public:
  int maxDepth(Node *root) {
    if (NULL == root) {
      return 0;
    }
    std::stack<pair<Node*, int>> mystack;    //深度遍历特点
    mystack.push(pair<Node*, int>(root, 1)); //从上到下传递深度
    int max_depth = 0; //统计从root节点到leaf节点之间最大距离

    while (!mystack.empty()) {
      Node *tempNode = mystack.top().first;
      int depth = mystack.top().second; //出栈 
      mystack.pop();

      //从左到右遍历,空节点不入栈
      for (Node *it : tempNode->children) {
        if (it) {
          mystack.push(pair<Node *, int>(it, depth + 1));
        }
      }
      //if (tempNode->children.size() ==0 ) { 添加 size 判断增加20ms的时间
        max_depth = max(max_depth, depth);
      //}
     
    }
    return max_depth;
  }
};

Java递归版本

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

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