LeetCode 0589 N-ary Tree Preorder Traversal【N叉树，DFS】

Given an n-ary tree, return the preorder traversal of its nodes’ values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Follow up:

Recursive solution is trivial, could you do it iteratively?

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]

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: [1,2,3,6,7,11,14,4,8,12,5,9,13,10

Constraints:

• The height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

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题意

N叉树前序遍历

思路1

• 递归实现

代码1

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

    Node() {}

    Node(int _val) {
        val = _val;
    }

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

class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> ans;
        if(root != NULL)
            dfs(root, ans);
        return ans;
    }
    
    void dfs(Node *node, vector<int> &ans){
        if(node == NULL) return;
        ans.push_back(node->val);
        for(const auto &item : node->children)
            dfs(item, ans);
    }
};