[Swift]LeetCode257. 二叉树的所有路径 | Binary Tree Paths

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

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

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

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

---

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。

示例:

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

---

16ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func binaryTreePaths(_ root: TreeNode?) -> [String] {
        var list:[String] = [String]()
        recuesive(root,&list,String())
        return list
    }
    func recuesive(_ root:TreeNode?,_ list:inout [String],_ str:String)
    {
        if root == nil {return}
        var strNew:String = str
        var strRoot:String = String(root!.val)
        if root?.left == nil && root?.right == nil
        {
            strNew = strNew + strRoot
            list.append(strNew)
            return
        }
        strRoot = strNew + strRoot + "->"
        recuesive(root?.left,&list,strRoot)
        recuesive(root?.right,&list,strRoot)
    }
}

---

16ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func binaryTreePaths(_ root: TreeNode?) -> [String] {
        guard let root = root else {
            return []
        }
        var result = [String]()
        binaryTreePathsDFS(root, "", &result)
        return result
    }
    
    func binaryTreePathsDFS(_ root: TreeNode, _ out: String, _ result: inout [String]) {
        if root.left == nil && root.right == nil {
            result.append(out + String(root.val))
        }
        
        if root.left != nil {
            binaryTreePathsDFS(root.left!, out + String(root.val) + "->", &result)
        }
        if root.right != nil {
            binaryTreePathsDFS(root.right!, out + String(root.val) + "->", &result)
        }
        
    }
}

---

16ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func binaryTreePaths(_ root: TreeNode?) -> [String] {
        var ans = [String]()
        binaryTreePaths(root, "", &ans)
        return ans
    }
    
    func binaryTreePaths(_ node: TreeNode?, _ path: String, _ ans: inout [String]) {
        guard let node = node else { return }
        
        let path = path + String(node.val)
        
        if node.left == nil && node.right == nil {
            ans.append(path)
            return 
        }
        
        binaryTreePaths(node.left, path + "->", &ans)
        binaryTreePaths(node.right, path + "->", &ans)
    }
}