Program to count leaf nodes in a binary tree

getLeafCount(node)
1) If node is NULL then return 0.
2) Else If left and right child nodes are NULL return 1.
3) Else recursively calculate leaf count of the tree using below formula.
    Leaf count of a tree = Leaf count of left subtree + 
                                 Leaf count of right subtree

Implementation:

#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; /* Function to get the count of leaf nodes in a binary tree*/ unsigned int getLeafCount(struct node* node) { if(node == NULL) return 0; if(node->left == NULL && node->right==NULL) return 1; else return getLeafCount(node->left)+getLeafCount(node->right); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } /*Driver program to test above functions*/ int main() { /*create a tree*/ struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /*get leaf count of the above created tree*/ printf("Leaf count of the tree is %d", getLeafCount(root)); getchar(); return 0; }

Time Complexity : O(n)
Proof:
T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of root and n-k-1 on the other side.
Let’s do analysis of boundary conditions
Case 1: Skewed tree (One of the subtrees is empty and other subtree is non-empty )
k is 0 in this case.
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c
…………………………………………
………………………………………….
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0) + (n)c
Value of T(0) will be some constant say d. (traversing a empty tree will take some constants time)
T(n) = n(c+d)
T(n) = (-)(n) (Theta of n)
Case 2: Both left and right subtrees have equal number of nodes.
T(n) = 2T(|_n/2_|) + c
This recursive function is in the standard form (T(n) = aT(n/b) + (-)(n) ) for master method http://en.wikipedia.org/wiki/Master_theorem. If we solve it by master method we get (-)(n)
Auxiliary Space : If we don’t consider size of stack for function calls then O(1) otherwise O(n).