How to display all descendant Vectors in Family Tree

[kingkush]

I am unsure how to go about solving this. I am almost certain solving this recursively is the way to go but I'm not quite sure how to go about it. I have a structure with a string and vector of said structure. The parent is the name and the vector are the children.
Structure Code:

struct TreeNode
{
    string data;
    vector<TreeNode *> children;
    TreeNode(string data)
    {
        this -> data = data;
    }
};

what i have which isnt much:

void ListDescendants(TreeNode * parentPtr)
{
    TreeNode *childPtr = nullptr;
    if(parentPtr -> children.size() > 0)
    {
        for(int i = 0; i < parentPtr -> children.size(); i++)
        {
            cout << parentPtr -> children[i] << endl;
            childPtr = parentPtr -> children[i];
        }
        ListDescendants(childPtr);
    }
    else if(parentPtr -> children.size() == 0)
    {
       cout << parentPtr -> data << " has no descendants. " << endl;
    }
}       {
            parentPtr = parentPtr -> children[i];
            ListDescendants(parentPtr);
        }
    }
    else if(parentPtr -> children.size() == 0)
    {
       cout << parentPtr -> data << " has no descendants. " << endl;
    }
}

I know i need a for loop but i cant seem to figure out the logic. I've been trying to draw pictures to help visualize it but to no avail. Can anyone help send me in the right direction?

[JLBorges]

struct TreeNode
{
    // ...
    std::string data;
    std::vector< TreeNode* > children;
};

void list_node( const TreeNode* p, int depth = 0 )
{ if(p) std::cout << std::string( depth*2, ' ' ) << p->data << '\n' ; }

void list_subtree( const TreeNode* p, int depth = 0 )
{
    list_node( p, depth ) ;
    if(p) for( const TreeNode* pchild : p->children ) list_subtree( pchild, depth+1 ) ;
}

http://coliru.stacked-crooked.com/a/864327f8f53878c8

[kingkush]

Wow that was a very thorough answer! Thank You! Could you help explain the depth part? I believe that was what was causing me issues in my logic. Thanks again!!

[JLBorges]

> Could you help explain the depth part?
> I believe that was what was causing me issues in my logic.

The depth is the depth of recursion (the level of the node in the sub-tree). Each time we descend one level down the tree, we pass the depth as current depth + 1
for( const TreeNode* pchild : p->children ) list_subtree( pchild, depth+1 ) ;
The depth of the node is then used to give the output a tree-like look, with the data of the node indented by the depth of the node; so that we get an output like this:

aaaa
  bbbb
    eeee
      pppp
      qqqq
      rrrr
        tttt
        uuuu
        vvvv
        wwww
      ssss
    ffff
    gggg
    hhhh
  cccc
    iiii
    jjjj
    kkkk
  dddd
    mmmm
    nnnn
    oooo

http://coliru.stacked-crooked.com/a/864327f8f53878c8

To generate a list all the nodes in the sub-tree, without any indentation to indicate the level, keeping track of the depth is not required.

void list_subtree( const TreeNode* p )
{
    if(p)
    {
        std::cout << p->data << '\n' ;
        for( const TreeNode* pchild : p->children ) list_subtree(pchild) ;
    }
}

In that case, we would get a flat list like this:

aaaa
bbbb
eeee
pppp
qqqq
rrrr
tttt
uuuu
vvvv
wwww
ssss
ffff
gggg
hhhh
cccc
iiii
jjjj
kkkk
dddd
mmmm
nnnn
oooo

http://coliru.stacked-crooked.com/a/dff47bcc7434f304

[kingkush]

Ahhh that makes sense. Thank you for the clarification(: