[数据结构与算法]Leetcode 589.N叉树的前序遍历

原题链接： Leetcode 589. N-ary Tree Preorder Traversal

关键词： 树

Given the root of 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)

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 number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

方法一：递归
如果为空，直接返回；
现将当前顶点放入结果，然后递归操作它的孩子

// c++ 做法

/*
// 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 getRes(const Node* root, vector<int> & res) {
        if (root == nullptr) {
            return;
        }
        res.emplace_back(root->val);
        for (auto & ch : root->children) {
            getRes(ch, res);
        }
    }

    // 前序遍历
    vector<int> preorder(Node* root) {
        vector<int> res;
        getRes(root, res);
        return res;
    }
};

# python做法

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def preorder(self, root: 'Node') -> List[int]:
        stack, ans = [root], []
        while stack:
            # 取出栈顶
            node = stack.pop()
            if node:
                ans.append(node.val)
                # pop 每次弹出最后的元素, 需要用[::-1]变为逆序，从最左孩子开始弹出
                for child in node.children[::-1]
                    stack.append(child)
        return ans