A node of binary search tree is given. Write a code which will left rotate the tree about given node.
SOLUTION
void left_rotate ( tree * x){
tree * y = x->right;
tree *z = x->parent;
x->right = y->left;
y->left->parent = x;
y->parent = z;
if ( z == NULL)
root = y;
else {
if ( x == z->left)
z->left = y;
else
z->right = y;
}
y->left = x;
x->parent = y;
}
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4 comments:
v nid d prnt f d gvn nde 2 slv dis prblm.
@above:
then find it out. Well, u may assume that parent pointer is there in tree structure.
tree *z = x->parent;
so u r having three pointers in ur node(left, right, parent) rather than traditional two (left and right) pointer.....
@above:
yes, i've already mentioned it in 2nd post.
Even if it is not there, we can search it out ,but this will make left rotate linear in time.
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