LeetCode——Maximum Depth of N-ary Tree（559）

最新推荐文章于 2025-11-15 21:38:30 发布

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本文探讨了如何在N叉树中找到最大深度的问题，通过遍历树结构并记录每个节点的深度，最终确定从根节点到最远叶节点的最长路径上的节点数，即为树的最大深度。

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.

For example, given a 3-ary tree:

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.

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

static const auto __ = [](){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return nullptr;
}();

class Solution {
public:
    int maxDepth(Node* root) {
        if(root==NULL)
            return 0;
        int max = 0, curr = 0;
        vector<Node*> stack;
        //stack.reserve(5001);
        vector<int> int_stack;
        //int_stack.reserve(5001);
        Node* temp;
        stack.push_back(root);
        int_stack.push_back(1);
        while(!stack.empty())
        {
            temp = stack.back();
            stack.pop_back();
            curr = int_stack.back();
            int_stack.pop_back();
            if(temp->children.size()==0)
                if(max < curr)
                    max = curr;
            for(int i=0;i<temp->children.size();++i)
            {
                stack.push_back(temp->children[i]);
                int_stack.push_back(curr+1);
            }
        }
        return max;
    }
};