Formatted question description: https://leetcode.ca/all/590.html
590. N-ary Tree Postorder Traversal (Easy)
Given an n-ary tree, return the postorder traversal of its nodes' values.
For example, given a 3-ary
tree:
Return its postorder traversal as: [5,6,3,2,4,1]
.
Note:
Recursive solution is trivial, could you do it iteratively?
Related Topics:
Tree
Similar Questions:
- Binary Tree Postorder Traversal (Hard)
- N-ary Tree Level Order Traversal (Easy)
- N-ary Tree Preorder Traversal (Easy)
Solution 1. Recursive
// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
private:
vector<int> ans;
void rec(Node *root) {
if (!root) return;
for (auto ch : root->children) rec(ch);
ans.push_back(root->val);
}
public:
vector<int> postorder(Node* root) {
rec(root);
return ans;
}
};
Solution 2. Iterative
Preorder traverse the mirrored tree and reverse the result!
// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/
// Time: O(N)
// Space: O(logN)
class Solution {
public:
vector<int> postorder(Node* root) {
if (!root) return {};
stack<Node*> s;
s.push(root);
vector<int> ans;
while (s.size()) {
root = s.top();
s.pop();
ans.push_back(root->val);
for (auto c : root->children) s.push(c);
}
reverse(ans.begin(), ans.end());
return ans;
}
};
Java
-
/* // Definition for a Node. class Node { public int val; public List<Node> children; public Node() {} public Node(int _val) { val = _val; } public Node(int _val, List<Node> _children) { val = _val; children = _children; } }; */ class Solution { public List<Integer> postorder(Node root) { List<Integer> postorder = new ArrayList<Integer>(); Stack<Node> stack1 = new Stack<Node>(); Stack<Node> stack2 = new Stack<Node>(); if (root != null) stack1.push(root); while (!stack1.isEmpty()) { Node temp = stack1.pop(); stack2.push(temp); List<Node> children = temp.children; int size = children.size(); for (int i = 0; i < size; i++) stack1.add(children.get(i)); } while (!stack2.isEmpty()) postorder.add(stack2.pop().val); return postorder; } }
-
// OJ: https://leetcode.com/problems/n-ary-tree-postorder-traversal/ // Time: O(N) // Space: O(logN) class Solution { private: vector<int> ans; void rec(Node *root) { if (!root) return; for (auto ch : root->children) rec(ch); ans.push_back(root->val); } public: vector<int> postorder(Node* root) { rec(root); return ans; } };
-
""" # Definition for a Node. class Node(object): def __init__(self, val, children): self.val = val self.children = children """ class Solution(object): def postorder(self, root): """ :type root: Node :rtype: List[int] """ res = [] if not root: return res for child in root.children: res.extend(self.postorder(child)) res.append(root.val) return res