【二叉树】N叉树的前序遍历

N叉树的后序遍历

1、题目

给定一个 N 叉树，返回其节点值的 前序遍历 。

N 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。

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

 

2、思路解析

1）递归

/*
// 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> ans;
    void dfs(Node* root)
    {
        if(!root){
            return;
        }
        ans.push_back(root->val);
        int size = root->children.size();
        for(int i = 0; i < size; ++i){
            if(root->children[i]){
                dfs(root->children[i]);
            }
        }
    }
    vector<int> preorder(Node* root) {
        dfs(root);
        return ans;
    }
};

2）迭代

/*
// 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;
        stack<Node*> s;

        if(!root) return ans;

        s.push(root);
        while(!s.empty())
        {
            Node* node = s.top();
            s.pop();
            if(node){
                ans.emplace_back(node->val);
            }
            else {
                continue;
            }

            if(!node->children.empty()){
                int size = node->children.size();
                for(int i = size - 1; i >= 0; --i){
                    Node* n = node->children[i];
                    if(n){
                        s.push(n);
                    }
                }
            }
        }
        return ans;
    }
};