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  • 590. N-ary Tree Postorder Traversal

    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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10201331.html
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