Real Networks Interview Question for Software Engineer / Developers






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of 0 vote

public void inorder(Tree node){
if(node.left!=null)
inorder(node.left)
system.out.println(node.data);
if(node.right!=null)
inorder(node.right);

}

- Anonymous August 31, 2010 | Flag Reply
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0
of 0 votes

perfectly correct

- scarlet September 15, 2010 | Flag
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0
of 0 vote

inorder(ptr)
struct node *ptr { if(ptr!=NULL){
inorder(ptr->lchild)
printf("%c",ptr->data)
inorder(ptr->rchild)}}

- prashant September 22, 2010 | Flag Reply
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0
of 0 vote

Both above codes are of preorder traveral.
left subtree -> Node -> righ subtree

For Inorder traversal, the order is
Node -> left subtree -> righ subtree

This one should work. Please comment.

public void inorder(Tree node)
{
	if(node == null) return;
	system.out.println(node.data);
	inorder(node.left);
	inorder(node.right);
}

- DG August 12, 2011 | Flag Reply


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