C#LeetCode刷题之#257-二叉树的所有路径（Binary Tree Paths）

问题

该文章的最新版本已迁移至个人博客【比特飞】，单击链接 https://www.byteflying.com/archives/4082 访问。

给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

输入:

1

 /   \

2     3

 \

  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

---

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Input:

1

 /   \

2     3

 \

  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

---

示例

该文章的最新版本已迁移至个人博客【比特飞】，单击链接 https://www.byteflying.com/archives/4082 访问。

public class Program {

    public static void Main(string[] args) {
        var root = new TreeNode(1) {
            left = new TreeNode(2) {
                right = new TreeNode(5)
            },
            right = new TreeNode(3)
        };

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

        root.right.right = new TreeNode(4);

        res = BinaryTreePaths2(root);
        ShowArray(res);

        Console.ReadKey();
    }

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

    public static IList<string> BinaryTreePaths(TreeNode root) {
        var list = new List<string>();
        if(root != null) {
            var left = BinaryTreePaths(root.left);
            var right = BinaryTreePaths(root.right);
            var current = root.val.ToString();
            if(left.Count == 0 && right.Count == 0) {
                list.Add(current);
            } else {
                if(left.Count != 0) {
                    foreach(var str in left) {
                        list.Add(current + "->" + str);
                    }
                }
                if(right.Count != 0) {
                    foreach(var str in right) {
                        list.Add(current + "->" + str);
                    }
                }
            }
        }
        return list;
    }

    public static IList<string> BinaryTreePaths2(TreeNode root) {
        var res = new List<string>();
        if(root == null) return res;
        TreePaths(res, root, root.val + "");
        return res;
    }

    public static void TreePaths(IList<string> res, TreeNode root, string path) {
        if(root.left == null && root.right == null)
            //若左右子节点都为空，本次路径结束，直接添加到 List
            res.Add(path);
        if(root.left != null)
            //若左子节点不为空，则将左子节点拼接之后参与下次递归
            TreePaths(res, root.left, path + "->" + root.left.val);
        if(root.right != null)
            //若右子节点不为空，则将右子节点拼接之后参与下次递归
            TreePaths(res, root.right, path + "->" + root.right.val);
    }

    public class TreeNode {
        public int val;
        public TreeNode left;
        public TreeNode right;
        public TreeNode(int x) { val = x; }
    }

}

以上给出2种算法实现，以下是这个案例的输出结果：

该文章的最新版本已迁移至个人博客【比特飞】，单击链接 https://www.byteflying.com/archives/4082 访问。

1->2->5 1->3
1->2->5 1->3->4

---

分析：

显而易见，以上2种算法的时间复杂度均为：  。