589.N-Ary Tree Preorder Traversal

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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]

我的解法

解题思路

递归

前序遍历一个树，树的子节点可能有多个，使用递归的方法

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/*
// 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:
    void preorder_traverse(vector<int>& nums, Node* tree_node){
        nums.push_back(tree_node -> val);
        for (auto iter : tree_node -> children){
            preorder_traverse(nums, iter);
        }
    }
    vector<int> preorder(Node* root) {
        vector<int> nums;
        if (root == NULL)
            return nums;
        preorder_traverse(nums, root);
        return nums;
    }
};

Runtime: 60 ms, faster than 85.95% of C++ online submissions for N-ary Tree Preorder Traversal.
Memory Usage: 56.7 MB, less than 10.53% of C++ online submissions for N-ary Tree Preorder Traversal.

迭代

迭代的方法：

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/*
// 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> nums;
        if (root == NULL) return nums;
        stack<Node*> s;
        s.push(root);
        while (!s.empty()){
            Node* tmp = s.top();
            s.pop();
            nums.push_back(tmp -> val);
            for (int i = tmp -> children.size() - 1; i >= 0; i--)
                s.push(tmp -> children[i]);
        }
        return nums;
    }
};

Runtime: 60 ms, faster than 85.95% of C++ online submissions for N-ary Tree Preorder Traversal.
Memory Usage: 56.5 MB, less than 10.53% of C++ online submissions for N-ary Tree Preorder Traversal.