CareerCup

maximum(node n):
  if n==null:
    return INTEGER.MINIMUM
  return max(n->data, maximum(n->left), maximum(n->right)

if its a BST, then we can move to the rightmost element and find the max value...otherwise i guess it takes O(n) complexity to find the max value in a binary tree(simply traverse the entire tree and find the max value)

This is the code I had written

node *max=root; //make it global;
findmax(root);
printf("Maximum val:%d",max->value);

void findmax(node *cur)
{
 if(cur->val>max->val)
   max=cur;
 findmax(cur->left);
 findmax(cur->right);
}

How about if we kept a seperate list which contains the max values. before you add or remove an element from the tree compare it to the first element on the list.

Since it is binary tree (and not binary search tree), pick you favourite traversal, and modify it to keep track of max etc.

int max(struct node* node) {
if(node==NULL) return -1;
int max=node->data, temp=-1;
if(node->left != NULL) { if(max < (temp=max(node->left))) max=temp;}
if(node->right != NULL) { if(max < (temp=max(node->right))) max=temp;}
return max; }

public static int findMax(Node start){
if(start == null)
return 0 ;
if(start.left == null && start.right == null)
return start.value;
if(start.left == null)
return Max(start.value , findMax(start.right));
if(start.right == null)
return Max(start.value , findMax(start.left));
return Max(start.value , Max(findMax(start.left), findMax(start.right)));

}


static int Max(int val1 , int val2){
if(val1 > val2)
return val1 ;
return val2;
}

int findmax (Node *root, int *maxm)
{
if (root)
{
findmax(root->left, maxm);
*maxm = max(root->data, *maxm);
findmax(root->right, maxm);
}
}

I ain't sure but Java doesn't have pointers, how do you plan to build a Binary tree in java?