1

The structure of the node is below. 

struct node
{
   int data;
   int noofchilds;
   node *child[n];
   node *parent;
 };

I would appreciate both recursive and non-recursive approaches.

Peter
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3 Answers3

1

Non-recursive version:

struct node {
    struct node *parent;
    unsigned nchild;
    struct node *child[XXX];
    int data;
    };

void deltree(struct node *np)
{
struct node *par;

while (np) {
        /* if this node has any children, start by
        ** "descending" to the highest numbered child and kill that first.
        */
        if (np->nchild--) {
                np = np->child[np->nchild];
                continue;
                }
        /* when we arrive here, *np has no more children left,
        ** so kill it, and step up to its parent
        */
        par = node->parent;
        // if np->child was obtained via malloc() uncomment next line
        // free (np->child);
        free (np);
        np = par;
        }
return;
}
wildplasser
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0

Remove the node and recursively remove its children.

If you have to remove the complete tree (as your question seems to be), the parent pointer does not matter (and is removed with the removal of the node itself).

Michel Keijzers
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  • yes but if i want to do it iteratively i.e without using a stack or queue, then what could be the approach , i think parent pointer might benefit us that time – Peter Feb 23 '12 at 13:56
  • Normally you start at the root since that is the only node you mostly have direct access too ... and in that case the parent is of no use since you only need to go 'downwards' ... I'm wondering if a iterative solution would help in this. – Michel Keijzers Feb 23 '12 at 13:59
  • node = root; do { if(node->noofchild){ node = node->child[noofchild-1]; else { node->parent->noofchild --; curr = node->parent; free(node); node = curr; } }while(node->parent || (node==root && node)) – Peter Feb 23 '12 at 14:23
0

An iterative algorithm:

Start at the parent node.

Then do the following actions as long as possible:
   if the current node has one or more children: 
      set one of the children nodes as the (next) current node
   else
      delete the current node and use its parent as the (next) current node.
      if the current node was the root node (which has no parent), stop.
jofel
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  • we cannot stop if current node is root because then only subtree correspoding to only 1 child will be true , but i think my algo below similar to yours will do – Peter Feb 23 '12 at 14:11
  • node = root; do { if(node->noofchild){ node = node->child[noofchild-1]; else { node->parent->noofchild --; curr = node->parent; free(node); node = curr; } }while(node->parent || (node==root && node)) – Peter Feb 23 '12 at 14:15
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    You are right. My algorithm is only fast draft... Your algorithm seems to have some problems, for example if the tree is empty. I would use node = root; while(node) { if(node->noofchild){ node->noofchild--; node = node->child[node->noofchild]; } else { curr = node->parent; free(node); node = curr; } } – jofel Feb 23 '12 at 14:31
  • we can make a check in starting in tree is empty and returning there itself – Peter Mar 11 '12 at 08:05