257. Binary Tree Paths

LeetCode easy original: C# #csharp #easy #leetcode #recursion #string #tree #two-pointers
Văn bản bài toán được dịch từ tiếng Nga theo ngôn ngữ giao diện. Mã không thay đổi.
LeetCode Static

given корневое cây, return все пути от корня до листа в любом порядке.

Лист — это узел без детей.

Ví dụ:

Input: root = [1,2,3,null,5]

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

C# lời giải

đã khớp/gốc 
public class Solution {
    public void ConstructPaths(TreeNode root, string path, IList<string> paths) {
        if (root != null) {
            path += root.val;
            if (root.left == null && root.right == null) {
                paths.Add(path);
            } else {
                path += "->";
                ConstructPaths(root.left, path, paths);
                ConstructPaths(root.right, path, paths);
            }
        }
    }
    public IList<string> BinaryTreePaths(TreeNode root) {
        IList<string> paths = new List<string>();
        ConstructPaths(root, "", paths);
        return paths;
    }
}

C++ lời giải

bản nháp tự động, xem lại trước khi gửi 
#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public void ConstructPaths(TreeNode root, string path, vector<string> paths) {
        if (root != null) {
            path += root.val;
            if (root.left == null && root.right == null) {
                paths.push_back(path);
            } else {
                path += "->";
                ConstructPaths(root.left, path, paths);
                ConstructPaths(root.right, path, paths);
            }
        }
    }
    public vector<string> BinaryTreePaths(TreeNode root) {
        vector<string> paths = new List<string>();
        ConstructPaths(root, "", paths);
        return paths;
    }
}

Java lời giải

đã khớp/gốc 
class Solution {
    public void construct_paths(TreeNode root, String path, LinkedList<String> paths) {
        if (root != null) {
            path += Integer.toString(root.val);
            if (root.left == null && root.right == null) {
                paths.add(path);
            } else {
                path += "->";
                construct_paths(root.left, path, paths);
                construct_paths(root.right, path, paths);
            }
        }
    }

    public List<String> binaryTreePaths(TreeNode root) {
        LinkedList<String> paths = new LinkedList<>();
        construct_paths(root, "", paths);
        return paths;
    }
}

JavaScript lời giải

đã khớp/gốc 
class Solution {
    constructPaths(root, path, paths) {
        if (root !== null) {
            path += root.val
            if (root.left === null && root.right === null) {
                paths.push(path)
            } else {
                path += "->"
                this.constructPaths(root.left, path, paths)
                this.constructPaths(root.right, path, paths)
            }
        }
    }

    binaryTreePaths(root) {
        const paths = []
        this.constructPaths(root, "", paths)
        return paths
    }
}

Python lời giải

đã khớp/gốc 
class Solution:
    def construct_paths(self, root: TreeNode, path: str, paths: List[str]):
        if root:
            path += str(root.val)
            if not root.left and not root.right:
                paths.append(path)
            else:
                path += "->"
                self.construct_paths(root.left, path, paths)
                self.construct_paths(root.right, path, paths)

    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        paths = []
        self.construct_paths(root, "", paths)
        return paths

Go lời giải

đã khớp/gốc 
func constructPaths(root *TreeNode, path string, paths *[]string) {
    if root != nil {
        path += fmt.Sprintf("%d", root.Val)
        if root.Left == nil && root.Right == nil {
            *paths = append(*paths, path)
        } else {
            path += "->"
            constructPaths(root.Left, path, paths)
            constructPaths(root.Right, path, paths)
        }
    }
}

func binaryTreePaths(root *TreeNode) []string {
    var paths []string
    constructPaths(root, "", &paths)
    return paths
}

Algorithm

1️⃣

Если текущий узел не является null, добавьте его значение к текущему пути.

Если текущий узел является листом (не имеет дочерних узлов), добавьте текущий путь в список путей.

Если текущий узел не является листом, добавьте "->" к текущему пути и рекурсивно вызовите функцию для левого и правого дочерних узлов.

2️⃣

Начните с корневого узла, пустого пути и пустого списка путей.

3️⃣

return список всех путей от корня до листа.

😎

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