AnwingWu
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bool findLargest(Node* t, int n, int& size, int& value)
{
int rightCount = 0;
if (t->right == NULL && n == 1)
{
value = t->value;
return true;
}
if (t->right != NULL)
if (findLargest(t->right, n, rightCount, value))
return true;
if (n == rightCount + 1)
return true;
if (t->left == NULL)
{
size = rightCount + 1;
return false;
}
int leftCount = 0;
if (findLargest(t->left, n - rightCount - 1, leftCount, value))
return true;
size = leftCount + rightCount + 1;
return false;
}
int findLargest(Tree* t, int n)
{
int size = 0, value = 0;
findLargest(t->root, n, size, value);
return value;
}
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a. you must use m + n - 2 steps to get to the destination, of which m - 1 is go right. So the total number of path is C(m-1, m + n - 2). C is the combinatiuon number.
- AnwingWu July 14, 2010b. Use dynamical programming, T.C. = O(mn), S.C. = O(mn)