590. N-ary Tree Postorder Traversal*
590. N-ary Tree Postorder Traversal*
https://leetcode.com/problems/n-ary-tree-postorder-traversal/
题目描述
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-arytree is less than or equal to 1000 - The total number of nodes is between
[0, 10^4]
C++ 实现 1
递归的方式.
/*
// 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) {if (!root) return {};vector<int> res;for (auto &c : root->children) {auto tmp = postorder(c);std::copy(tmp.begin(), tmp.end(), back_inserter(res));}res.push_back(root->val);return res;}
};
C++ 实现 2
迭代的方式. 迭代的方式可以仔细研究一下, children 按顺序入栈, 最后 reverse res.
class Solution {public:vector<int> postorder(Node* root) {if (!root) return {};vector<int> res;stack<Node*> st;st.push(root);while (!st.empty()) {auto r = st.top();st.pop();res.push_back(r->val);for (auto &c : r->children) st.push(c);}std::reverse(res.begin(), res.end());return res;}
};
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