C#LeetCode刷题之#590-N叉树的后序遍历(N-ary Tree Postorder Traversal)

C#LeetCode刷题之#590-N叉树的后序遍历(N-ary Tree Postorder Traversal)

问题

给定一个 N 叉树,返回其节点值的后序遍历

例如,给定一个 3叉树 :

C#LeetCode刷题之#590-N叉树的后序遍历(N-ary Tree Postorder Traversal)

返回其后序遍历: [5,6,3,2,4,1].

说明: 递归法很简单,你可以使用迭代法完成此题吗?

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?


示例

public class Program {

    public static void Main(string[] args) {
        var root = new Node(1,
                new List<Node> {
            new Node(3, new List<Node> {
                new Node(5, null),
                new Node(6, null)
            }),
            new Node(2, null),
            new Node(4, null)
            });

        var res = Postorder(root);
        ShowArray(res);

        res = Postorder2(root);
        ShowArray(res);

        Console.ReadKey();
    }

    private static void ShowArray(IList<int> array) {
        foreach(var num in array) {
            Console.Write($"{num} ");
        }
        Console.WriteLine();
    }

    public static IList<int> Postorder(Node root) {
        var list = new List<int>();
        Post(root, ref list);
        return list;
    }

    public static void Post(Node root, ref List<int> list) {
        if(root == null) return;
        if(root.children == null) {
            list.Add(root.val);
            return;
        }
        foreach(var child in root.children) {
            Post(child, ref list);
        }
        list.Add(root.val);
        return;
    }

    public static IList<int> Postorder2(Node root) {
        if(root == null) return new List<int>();
        var list = new List<int>();
        if(root.children != null) {
            foreach(var node in root.children) {
                list.AddRange(Postorder2(node));
            }
        }
        list.Add(root.val);
        return list;
    }

    public class Node {
        public int val;
        public IList<Node> children;
        public Node() { }
        public Node(int _val, IList<Node> _children) {
            val = _val;
            children = _children;
        }
    }

}

以上给出2种算法实现,以下是这个案例的输出结果:

5 6 3 2 4 1
5 6 3 2 4 1

分析:

显而易见,以上2种算法的时间复杂度均为: O(n)​ 。

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