589. N-ary Tree Preorder Traversal*

589. N-ary Tree Preorder Traversal*

​​https://leetcode.com/problems/n-ary-tree-preorder-traversal/​​

题目描述

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?

Example 1:

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

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: [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]​​

C++ 实现 1

递归版本.

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

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> res;
        res.push_back(root->val);
        for (auto &c : root->children) {
            auto vals = preorder(c);
            std::copy(vals.begin(), vals.end(), back_inserter(res));
        }
        return res;
    }
};

C++ 实现 2

迭代版本. 使用栈实现.

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

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> res;
        stack<Node*> st;
        st.push(root);
        while (!st.empty()) {
            auto r = st.top();
            st.pop();
            res.push_back(r->val);
            auto size = r->children.size();
            for (int i = size - 1; i >= 0; --i)
                st.push(r->children[i]);
        }
        return res;
    }
};