Answer:
Each time you insert a new node, call the function to adjust the sum.
This method has to be called each time we insert new node to the tree since the sum at all the
parent nodes from the newly inserted node changes when we insert the node.
// toSumTree method will convert the tree into sum tree.
int toSumTree(struct node *node)
{
if(node == NULL)
return 0;
// Store the old value
int old_val = node->data;
// Recursively call for left and right subtrees and store the sum as new value of this node
node->data = toSumTree(node->left) + toSumTree(node->right);
// Return the sum of values of nodes in left and right subtrees and
// old_value of this node
return node->data + old_val;
}
This has the complexity of O(n).
Explanation:
The answer is d torque because
Answer:
what do you need help with
Yes you do have to program it first.
JAVA programming was employed...
What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules
Solution:
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x1[i][j] = num.nextInt();
}
}// End of Array 1, x1.
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x2[i][j] = num.nextInt();
}
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:
x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
x1rules = true;
system.out.print(x1rules);
}
else
{
system.out.print(x1rules);
}//Conditional Statement
