Subtrees of height h in Binary tree - recursion

Question

How do we find number of subtrees of height 'h' in a binary tree.
Function is defined as
int subtree( node *root, int k);
where k is the specific height.

Answer 1

First, we recursively calculate the height of the tree as follows:
If the tree is empty, the height is 0.
If the tree is non-empty, the height is the maximum height of its children, plus 1.
In C (I assume OP is using C based on response), this looks like
typedef struct Node {
Node* leftChild,
node* rightChild
} Node;
typedef Node* Tree;
unsigned int max(unsigned int a, unsigned int b) {
return a > b ? a : b;
}
unsigned int height(Tree tree) {
return tree ? 1 + max(height(tree->leftChild, tree->rightChild)) : 0;
}
Note that generally, Node will have some additional data. But that data isn't relevant here, so we don't include it (although it's easy enough to do so if you wish to).
Now, we want to modify the height function slightly. To do this, we define
typdef struct Result {
unsigned int height,
unsigned int count
} Result;
/**
* The returned .height should be the height of the tree.
* The returned .count should be the number of subtrees of tree
* with height k.
*/
Result resultCountChildren(Tree tree, unsigned int k) {
if (tree) {
Result leftResult = resultCountChildren(tree->left, k);
Result rightResult = resultCountChildren(tree->right, k);
unsigned int heightOfTree = 1 + max(leftResult.height, rightResult.height);
unsigned int count = leftResult.count + rightResult.count + (heightOfTree == k);
Result result = {
.height = heightOfTree,
.count = count
};
return result;
} else {
unsigned int height = 0;
unsigned int count = (0 == k);
Result result = {
.height = height,
.count = count
};
return result;
}
}
unsigned int count(Tree tree, unsigned int k) {
return resultCountChildren(tree).count;
}