美文网首页
二叉树 for iOS (Swift篇)

二叉树 for iOS (Swift篇)

作者: 程序H | 来源:发表于2017-09-26 11:43 被阅读128次

学习了二叉树的概念二叉树算法习题后, 结合学习的知识通过代码实现一些功能。

二叉树说白了是考验程序员的递归思维,通过自己调用自己来实现功能,废话不多说上代码。

创建二叉树类

创建BinaryTreeNode类,并分别创建value、leftNode、rightNode属性,导入数组创建二叉树

class BinaryTreeNode: NSObject {
   // 值
   var value:Int = 0
   // 左节点
   var leftNode:BinaryTreeNode?
   // 右节点
   var rightNode:BinaryTreeNode?
   
   /// 创建二叉树排序树节点
   /// 左节点值全部小于根节点,右节点值全部大于根节点
   ///
   /// - Parameter values: 数组
   /// - Returns: 二叉树根节点
   class func creatTree(values: [Int]) -> BinaryTreeNode {
        var root:BinaryTreeNode?
        for i in 0..<values.count {
            let value = values[i]
            root = BinaryTreeNode.addTreeNode(treeNode: root, value: value)
        }
        return root!
    }
    
   /// 向二叉排序树添加一个节点
   ///
   /// - Parameters:
   ///   - treeNode: 根节点
   ///   - value: 值
   /// - Returns: 根节点
   private class func addTreeNode(treeNode: BinaryTreeNode?, value:Int) -> BinaryTreeNode? {
        var treeNode = treeNode
        if treeNode == nil {
            treeNode = BinaryTreeNode()
            treeNode?.value = value
            print("node: \(value)")
        } else if value <= treeNode!.value {
            print("to left")
            // 值小于根节点
            treeNode?.leftNode = BinaryTreeNode.addTreeNode(treeNode: treeNode?.leftNode, value: value)
        } else {
            print("to right")
            // 值大于根节点
            treeNode?.rightNode = BinaryTreeNode.addTreeNode(treeNode: treeNode?.rightNode, value: value)
        }
        return treeNode
   }
}      

在合适的地方调用方法创建二叉树,笔者为了清晰的看到树的结构,通过笨方法打印了每层子节点:

let array = [1, 2, 10, 8, 3, 4, 5, 6, 7, 9]
let tree = BinaryTreeNode.creatTree(values: array)

print("<===============> 1层")
print(tree.value)

print("<===============> 2层")
print(tree.leftNode?.value)
print(tree.rightNode?.value)

print("<===============> 3层")
print(tree.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.value)

print("<===============> 4层")
print(tree.rightNode?.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.rightNode?.value)

print("<===============> 5层")
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.rightNode?.value)

print("<===============> 6层")
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.rightNode?.rightNode?.value)

print("<===============> 7层")
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.rightNode?.value)

print("<===============> 8层")
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.rightNode?.rightNode?.value)

print("<===============> 9层")
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.rightNode?.rightNode?.leftNode?.value)
print(tree.rightNode?.rightNode?.leftNode?.leftNode?.rightNode?.rightNode?.rightNode?.rightNode?.value)

打印结果
<===============> 1层
1
<===============> 2层
nil
Optional(2)
<===============> 3层
nil
Optional(10)
<===============> 4层
Optional(8)
nil
<===============> 5层
Optional(3)
Optional(9)
<===============> 6层
nil
Optional(4)
nil
nil
<===============> 7层
nil
Optional(5)
<===============> 8层
nil
Optional(6)
<===============> 9层
nil
Optional(7)

获取二叉树的深度(层数)

/// 二叉树的深度
///
/// - Parameter rootNode: 二叉树根节点
/// - Returns: 二叉树的深度
class func depthOfTree(rootNode: BinaryTreeNode?) -> Int {
    if rootNode == nil {
        return 0
    }
    if rootNode?.leftNode == nil && rootNode?.rightNode == nil {
        return 1
    }
    
    // 左子树深度
    let leftDepth = depthOfTree(rootNode: rootNode?.leftNode)
    // 右子树深度
    let rightDepth = depthOfTree(rootNode: rootNode?.rightNode)
    
    return max(leftDepth, rightDepth) + 1
}

在合适的地方调用:

print("二叉树深度\(BinaryTreeNode.depthOfTree(rootNode: tree))")
打印结果:
二叉树深度9

按层遍历

/// 二叉树中某个位置的节点(按层次遍历)
///
/// - Parameters:
///   - index: 按层次遍历的位置
///   - rootNode: 根节点
/// - Returns: tree
class func treeNodeAtIndex(index: Int, rootNode: BinaryTreeNode?) -> BinaryTreeNode? {
    var index = index
    if index == 0 && rootNode == nil {
        return nil
    }
    // 数组当成队列
    var queueArray = [BinaryTreeNode?]()
    // 压入根节点
    queueArray.append(rootNode!)
    while queueArray.count > 0 {
        let node = queueArray.first as? BinaryTreeNode
        if index == 0 {
            return node
        }
        // 弹出最前面的节点,返照队列先进先出的原则
        queueArray.remove(at: 0)
        // 节点移除,index减少
        index -= 1
        if node?.leftNode != nil {
            queueArray.append(node?.leftNode)
        }
        if node?.rightNode != nil {
            queueArray.append(node?.rightNode)
        }
    }
    return nil
}

在合适的地方调用:

let indexNode = BinaryTreeNode.treeNodeAtIndex(index: 2, rootNode: tree)
print(indexNode?.value)

打印结果:
Optional(10)

前、中、后遍历

/// 前序遍历
/// 先访问根,再遍历左子树,再遍历右子树,典型的递归思想
/// - Parameters:
///   - rootNode: 树
///   - handler: 回调数值
class func preOrderTraverseTree(rootNode: BinaryTreeNode? ,handler:@escaping (BinaryTreeNode) -> Void) {
    if rootNode != nil {
        handler(rootNode!)
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.leftNode, handler: handler)
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.rightNode, handler: handler)
    }
}

/// 中序遍历
/// 先遍历左子树,再访问根,再遍历右子树
/// - Parameters:
///   - rootNode: 树
///   - handler: 回调数值
class func inOrderTreverseTree(rootNode: BinaryTreeNode? ,handler:@escaping (BinaryTreeNode) -> Void) {
    if rootNode != nil {
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.leftNode, handler: handler)
        handler(rootNode!)
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.rightNode, handler: handler)
    }
}

/// 后序遍历
/// 先遍历左子树,再遍历右子树,在访问根
/// - Parameters:
///   - rootNode: 树
///   - handler: 回调数值
class func postOrderTreverseTree(rootNode: BinaryTreeNode? ,handler:@escaping (BinaryTreeNode) -> Void) {
    if rootNode != nil {
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.leftNode, handler: handler)
        BinaryTreeNode.preOrderTraverseTree(rootNode: rootNode?.rightNode, handler: handler)
        handler(rootNode!)
    }
}

在合适的地方调用:

// 前序遍历
var dataArr1 = [Int]()
BinaryTreeNode.preOrderTraverseTree(rootNode: tree) { (treeNode) in
    dataArr1.append(treeNode.value)
}
print(dataArr1)


// 中序遍历
var dataArr2 = [Int]()
BinaryTreeNode.inOrderTreverseTree(rootNode: tree) { (treeNode) in
    dataArr2.append(treeNode.value)
}
print(dataArr2)

// 后序遍历
var dataArr3 = [Int]()
BinaryTreeNode.postOrderTreverseTree(rootNode: tree) { (treeNode) in
    dataArr3.append(treeNode.value)
}
print(dataArr3)

打印结果:
// 前序遍历
[1, 2, 10, 8, 3, 4, 5, 6, 7, 9]
// 中序遍历
[1, 2, 10, 8, 3, 4, 5, 6, 7, 9]
// 后续遍历
[2, 10, 8, 3, 4, 5, 6, 7, 9, 1]

持续更新中~

相关文章

网友评论

      本文标题:二叉树 for iOS (Swift篇)

      本文链接:https://www.haomeiwen.com/subject/qulsextx.html