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:
```bash 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]
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;
}
};