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?

 

M1: recursive

time: O(n), space: O(height)

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> res = new ArrayList<>();
        postorder(root, res);
        return res;
    }
    
    public void postorder(Node root, List<Integer> res) {
        if(root == null) {
            return;
        }
        
        for(int i = 0; i < root.children.size(); i++) {
            postorder(root.children.get(i), res);
        }
        res.add(root.val);
    }
}

 

M2: iterative

time: O(n), space: O(n)　　-- depending on the tree structure, worst case: the stack can keep up the whole tree nodes

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        Stack<Node> s = new Stack<>();
        
        s.add(root);
        while(!s.isEmpty()) {
            Node tmp = s.pop();
            res.add(0, tmp.val);
            for(Node n : tmp.children) {
                if(n != null) {
                    s.push(n);
                }
            }
        }
        
        return res;
    }
}

 

二刷：

class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        LinkedList<Node> stack = new LinkedList<>();
        stack.offerFirst(root);
        while(!stack.isEmpty()) {
            Node cur = stack.pollFirst();
            res.add(0, cur.val);
            for(int i = 0; i < cur.children.size(); i++) {
                stack.offerFirst(cur.children.get(i));
            }
        }
        return res;
    }
}