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;
}
}