589.N-ary Tree Preorder Traversal

Tags: Tree Easy

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

---

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]

---

递归解法：

/*
// 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:
    vector<int> preorder(Node* root) {
        vector<int> res;
        if (!root) return res;

        preorder(root, res);

        return res;
    }

    void preorder(Node *root, vector<int> & res) {
        res.push_back(root -> val);
        for (auto e : root -> children) {
            if (e) preorder(e, res);
        }
    }
};

Runtime: 68 ms, faster than 93.71% 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.

非递归的写法，和二叉树基本一样，小小的区别就是之前二叉树入栈是先右再左，这里其实就是逆序遍历数组。

/*
// 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> res;
        if (!root) return res;

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

        return res;
    }
};

Runtime: 68 ms, faster than 93.71% 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.