559. Maximum Depth of N-ary Tree
阿新 • • 發佈:2018-10-14
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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.
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
#include<vector> #include <cstdlib> #include<iostream> using namespace std; // Definition for a Node. class Node { public: int val; vector<Node *> children; Node() {} Node(int _val, vector<Node *> _children) { val = _val; children = _children; } };class Solution { public: int maxDepth(Node *root) { if (root == NULL) return 0; int res = 1; for (int i = 0; i < root->children.size(); ++i) { res = max(res, 1 + maxDepth(root->children[i])); } return res; } }; int main() { Node*node5 = new Node(5, {}); Node *node6 = new Node(6, {}); Node *node3 = new Node(3, {node5, node6}); Node *node2 = new Node(2, {}); Node *node4 = new Node(4, {}); Node *node1 = new Node(1, {node3, node2, node4}); Solution solution; int res = solution.maxDepth(node1); std::cout << "res: " << res << endl; return 0; }
559. Maximum Depth of N-ary Tree