Question

我在操场文件中提取了问题，以使讨论更加容易。我的问题是，我需要找出树中具有最小子节点数的节点的所有值。我没有正确的递归。我的算法的结果为[1,9]，但应为：[1,3,9]。有什么想法吗？

import Foundation

class TreeNode{

    var children: [TreeNode]?
    var value: Int

    init( value: Int, children: [TreeNode]?){
        self.value = value
        self.children = children
    }
}

let node11 = TreeNode(value: 11, children: nil)
let node10 = TreeNode(value: 10, children: nil)
let node9 = TreeNode(value: 9, children: [node10, node11])
let node8 = TreeNode(value: 8, children: nil)
let node7 = TreeNode(value: 7, children: [node9])
let node6 = TreeNode(value: 6, children: [node8])
let node5 = TreeNode(value: 5, children:nil)
let node4 = TreeNode(value: 4, children: [node7])
let node3 = TreeNode(value: 3, children: [node5, node6])
let node2 = TreeNode(value: 2, children: [node4])
let node1 = TreeNode(value: 1, children: [node2,node3])
let node0 = TreeNode(value: 0, children: [node1])


func valueOfNodes(minNumberOfChildren: Int, node: TreeNode, tempResult: [Int])->[Int]{

    var result = tempResult

    guard let children = node.children else{
        return result
    }

    if children.count >= minNumberOfChildren{
        result.append(node.value)
    }

    for child in children{
        return valueOfNodes(minNumberOfChildren: minNumberOfChildren, node: child, tempResult: result)
    }

    return result
}


let result = valueOfNodes(minNumberOfChildren: 2, node: node0, tempResult: [])
print(result)

Answer 1

您始终只返回第一个孩子的结果：

def getTerm(self,value):
    catalog = api.portal.get_tool(name='portal_catalog')
    brains = catalog.searchResults({'portal_type':'Organization',
                                   'Title':value,
                                 })
    return SimpleTerm(title=brains[0]['Title'],
                      value=brains[0]['Title'],
                  )

相反，您需要将它们结合起来

for child in children{
    return valueOfNodes(minNumberOfChildren: minNumberOfChildren, node: child, tempResult: result)
}

这种方法终止了尾递归优化，但我认为您不能在此处使用它，因为父级的结果取决于所有子级的结果。

Swift：确定树中的节点

1 个答案: