590.N-ary Tree Postorder Traversal

Tags: Easy Tree

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

---

Given an n-ary tree, return the postorder 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: [5,6,3,2,4,1]

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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

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> postorder(Node* root) {
        vector<int> res;
        if (!root) return res;

        postorder(root, res);

        return res;
    }

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

Runtime: 68 ms, faster than 93.40% of C++ online submissions for N-ary Tree Postorder Traversal.
Memory Usage: 56.8 MB, less than 20.00% of C++ online submissions for N-ary Tree Postorder Traversal.

使用辅助栈的非递归解法：

/*
// 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> postorder(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 (auto e : p -> children) {
                if (e) s.push(e);
            }
        }
        reverse(res.begin(), res.end());

        return res;
    }
};

Runtime: 72 ms, faster than 91.74% of C++ online submissions for N-ary Tree Postorder Traversal.
Memory Usage: 56.7 MB, less than 20.00% of C++ online submissions for N-ary Tree Postorder Traversal.

比较耗时间的步骤是数组翻转的步骤。