show function of binary tree structure (a solution is given).

using DataStructures
tree = RBTree{Int}();
for k in 1:2:20
    push!(tree, k)
end
tree

The above code produces the following long output:

RBTree{Int64}(DataStructures.RBTreeNode{Int64}(false, 7, DataStructures.RBTreeNode{Int64}(false, 3, DataStructures.RBTreeNode{Int64}(false, 1, DataStructures.RBTreeNode{Int64}(false, nothing, nothing, nothing, 
nothing), DataStructures.RBTreeNode{Int64}(false, nothing, nothing, nothing, nothing), DataStructures.RBTreeNode{Int64}(#= circular reference @-2 =#)), 
……
DataStructures.RBTreeNode{Int64}(false, nothing, nothing, nothing, nothing), 10)

This is because the show function of binary tree is missing. So I developed the show function as follows:

Base.show(io::IO, tree::Union{RBTree,AVLTree,SplayTree}) = Base.show(io::IO, tree.root)

function Base.show(io::IO, root::Union{DataStructures.RBTreeNode,DataStructures.AVLTreeNode,DataStructures.SplayTreeNode})
    lowbit(x::Int) = Int(log2(x & (~x+1)))
    function printLevel(nodeList::Vector)
        nodeList1 = Any[]
        for (n,node) in enumerate(nodeList)
            n ==1 || (str *= string(repeat("·",lowbit(n-1))))
            if node == nothing
                append!(nodeList1, [nothing, nothing])
                str *= string("[,]")
                continue
            end
            data = node.leftChild == nothing ? string() : string(node.leftChild.data)
            str *= string('[', data)
            data = node.rightChild == nothing ? string() : string(node.rightChild.data)
            str *= string(',', data, ']')
            append!(nodeList1, [node.leftChild, node.rightChild])
        end
        return nodeList1
    end

    h = 8 # only show the first $h levels
    level = 0
    nodeList = [root]
    str = string(level, ": ", root.data)
    for i = 1:h  
        println(io,str)
        str = string(level + 1, ": ")
        nodeList = printLevel(nodeList)
        all(nodeList .== nothing) && return
        level += 1
    end
    level == h && println("......")
end

Then the above RBTree show like this:

0: 7    
1: [3,11]
2: [1,5][9,15]
3: [nothing,nothing][nothing,nothing]·[nothing,nothing][13,17]
4: [,][,]·[,][,]··[,][,]·[nothing,nothing][nothing,19]
5: [,][,]·[,][,]··[,][,]·[,][,]···[,][,]·[,][,]··[,][,]·[,][nothing,nothing]  

And the AVLTree show like this:

tree = AVLTree{Int}()
for k in 1:2:20
    push!(tree, k)
end
tree

0: 7    
1: [3,15]
2: [1,5][11,17]
3: [,][,]·[9,13][,19]

This function works with RBTree, AVLTree and SplayTree. The following is a brief description of the display:

1. The number of the tree level is displayed at the beginning of each line.
2. [a,b] have the same parent node.
3. [a,b][c,d] have the same grandparent node.
4. [a,b]·[c,d] have the same great-grandparent node.
5. [a,b]··[c,d] have the same great-great-grandparent node , and so on.

I hope a developer can check my code and pull it!

Start by reviewing the proposed Base.show methods for RBTree, AVLTree, and SplayTree and compare them with the existing tree display behavior. Confirm how each node type represents children and decide how the output should handle empty or deeper levels. Done means the three tree types display a readable structure without the long internal representation, with coverage for the examples in the issue.