Question

此serialize（）函数的错误在哪里

我在Swift中编写了一个deserialize（）函数，以从可选的Integer数组构造一个二叉树。我通过将输出与使用相同数组的手动方法构造的树的输出进行比较来测试此功能。他们看起来一样，这很好！

但是，当我运行另一个函数isSameTree（）时，该函数用于比较2棵树（我确定此函数正常工作），在deserialize（）的输出和manual方法的输出上，我得到不同的结果！

我认为deserialize（）不正确，但是我找不到错误！

//辅助代码

public class BinaryNode {
     public var value: Int
     public var left: BinaryNode?
     public var right: BinaryNode?
     public init(_ value: Int) {
         self.value = value
         self.left = nil
         self.right = nil
     }
 }

extension BinaryNode {
    public var description: String {
        return diagram(for: self)
    }
    private func diagram(for node: BinaryNode?,
                         _ top: String = "",
                         _ root: String = "",
                         _ bottom: String = "") -> String {
        guard let node = node else {
            return root + "nil\n"
        }
        if node.left == nil && node.right == nil {
            return root + "\(node.value)\n"
        }
        return diagram(for: node.right,
                       top + " ", top + "┌──", top + "│ ")
            + root + "\(node.value)\n"
            + diagram(for: node.left,
                      bottom + "│ ", bottom + "└──", bottom + " ")
    }
}

public func deserialize(_ array: inout [Int?]) -> BinaryNode? {
    guard !array.isEmpty, let value = array.removeFirst() else {
        return nil
    }
    let node = BinaryNode(value)
    node.left = deserialize(&array)
    node.right = deserialize(&array)
    return node
}

func isSameTree(_ p: BinaryNode?, _ q: BinaryNode?) -> Bool {
        guard let p = p else {
            return q == nil
        }
        guard let q = q else {
            return p == nil
        }

        if p.value != q.value {
            return false
        }

        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right)
    }


// Using deserialize to construct trees
var a1: [Int?] = [1,nil,2,3]
var a2: [Int?] = [1,nil,2,nil,3]
if let tree = deserialize(&a1) {
    print(tree.description)
}

if let tree = deserialize(&a2) {
    print(tree.description)
}

// Using manual to construct trees
let a3: BinaryNode = {
    let one = BinaryNode(1)
    let two = BinaryNode(2)
    let three = BinaryNode(3)
    one.right = two
    two.left = three
    return one
}()
print(a3.description)
let a4: BinaryNode = {
    let one = BinaryNode(1)
    let two = BinaryNode(2)
    let three = BinaryNode(3)
    one.right = two
    two.right = three
    return one
}()
print(a4.description)

// The print statements above show similar trees are constructed
// However, below results are not same

isSameTree(deserialize(&a1), deserialize(&a2)) // true <- this is wrong
isSameTree(a3, a4) // false <--- this is correct

Answer 1

您似乎忘记了deserialize(_:)对其参数的破坏性...请记住为什么需要&。

//Re-load a1 & a2...
a1 = [1,nil,2,3]
a2 = [1,nil,2,nil,3]
print(isSameTree(deserialize(&a1), deserialize(&a2))) //-> false
print(isSameTree(a3, a4)) //-> false

如何解决此二叉树反序列化问题？

1 个答案: