Question

如何在R中为给定数量的分支和深度构建N-ary树，例如深度为3的二叉树？

编辑：将来源问题与问题和答案分开。

Answer 1

我想提出解决方案，我用它来构建具有 leafAmn 分支因子的树数据结构。要在树中存储数据，请使用字段 myData 。还定义了打印树的内容的功能。此外，它还存在此任务的递归解决方案：R Tree With n Branches。

# function to build N-ary tree
makeTree <- function(depth, leafAmn)  
{
## leafAmn - branching factor
## depth   - depth of tree
library(data.tree)

myTree <- Node$new("root", myData = c(-1)) # root node
for (i in 1:depth) # loop by tree depth
{
    if (i == 1)
    # create a set of nodes with depth 1
    {
        chldArr1 <- matrix("", 1, leafAmn)
        for (j in 1:leafAmn)
        {
            # create children nodes
            myTree$AddChild(j, myData = c())
            # save links to children nodes to array 'chldArr1'
            # this array is used to generate tree without using recursion
            chldArr1[j] <- sprintf("myTree$children[[%d]]$", j)
        }
    }
    else
    # add childs at level 'i' to nodes at level 'i-1' using 
    # method AddChild
    {
        chldArr2 <- matrix("", 1, (leafAmn ^ i))
        k <- 1
        for (j in 1:(leafAmn ^ (i - 1)))
        {
            for (m in 1:leafAmn)
            {
                # create string which contains a command to execute
                # this command is used to add child to nodes at previous level
                commStr <- paste(chldArr1[j], sprintf("AddChild(%d, myData = c())", m), sep = "")
                eval(parse(text = commStr))
                print(commStr)
                # save command to array 'chldArr2'
                chldArr2[k] <- paste(chldArr1[j], sprintf("children[[%d]]$", m), sep = "")
                k <- k + 1
            }
        }
        chldArr1 <- chldArr2
    }
}

## Make a tree with depth of '3' and 2 branches from each node
myTree <- makeTree(3, 2)
print(myTree, "myData")

>     myTree <- makeTree(3, 2)
[1] "myTree$children[[1]]$AddChild(1, myData = c())"
[1] "myTree$children[[1]]$AddChild(2, myData = c())"
[1] "myTree$children[[2]]$AddChild(1, myData = c())"
[1] "myTree$children[[2]]$AddChild(2, myData = c())"
[1] "myTree$children[[1]]$children[[1]]$AddChild(1, myData = c())"
[1] "myTree$children[[1]]$children[[1]]$AddChild(2, myData = c())"
[1] "myTree$children[[1]]$children[[2]]$AddChild(1, myData = c())"
[1] "myTree$children[[1]]$children[[2]]$AddChild(2, myData = c())"
[1] "myTree$children[[2]]$children[[1]]$AddChild(1, myData = c())"
[1] "myTree$children[[2]]$children[[1]]$AddChild(2, myData = c())"
[1] "myTree$children[[2]]$children[[2]]$AddChild(1, myData = c())"
[1] "myTree$children[[2]]$children[[2]]$AddChild(2, myData = c())"
>     print(myTree, "myData")
       levelName myData
1  root              -1
2   ¦--1             NA
3   ¦   ¦--1         NA
4   ¦   ¦   ¦--1     NA
5   ¦   ¦   °--2     NA
6   ¦   °--2         NA
7   ¦       ¦--1     NA
8   ¦       °--2     NA
9   °--2             NA
10      ¦--1         NA
11      ¦   ¦--1     NA
12      ¦   °--2     NA
13      °--2         NA
14          ¦--1     NA
15          °--2     NA

Answer 2

好吧，很久以前，当Q-BASIC没有任何指针选项时，有一种方法可以表示没有指针的二叉树。这是一个简单的数学技巧。当您为root分配1时，它有两个子项 - 2,3。 “2”有两个子节点“4,5”，可以表示为“索引* 2”和“索引* 2 + 1”。因此，我们可以用数组表示二叉树。这是我的代码。

let x = a.filter { !b.contains($0) }

在R中建立N-ary树

2 个答案: