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C Program to Construct a Balanced Binary Tree using Sorted Array
#include <stdio.h> #include <stdlib.h> struct btnode int value; struct btnode *l; struct btnode *r; ; typedef struct btnode N; N* bst(int arr[], int first, int last); N* new(int val); void display(N *temp); int main() int arr[] = 10, 20, 30, 40, 60, 80, 90; N *root = (N*)malloc(sizeof(N)); int n = sizeof(arr) / sizeof(arr[0]), i; printf("Given sorted array is\n"); for (i = 0;i < n;i++) printf("%d\t", arr[i]); root = bst(arr, 0, n - 1); printf("\n The preorder traversal of binary search tree is as follows\n"); display(root); printf("\n"); return 0; /* To create a new node */ N* new(int val) N* node = (N*)malloc(sizeof(N)); node->value = val; node->l = NULL; node->r = NULL; return node; /* To create a balanced binary search tree */ N* bst(int arr[], int first, int last) int mid; N* temp = (N*)malloc(sizeof(N)); if (first > last) return NULL; mid = (first + last) / 2; temp = new(arr[mid]); temp->l = bst(arr, first, mid - 1); temp->r = bst(arr, mid + 1, last); return temp; /* To display the preorder */ void display(N *temp) printf("%d->", temp->value); if (temp->l != NULL) display(temp->l); if (temp->r != NULL) display(temp->r);
Output
Given sorted array is 10 20 30 40 60 80 90 The preorder traversal of binary search tree is as follows 40 - > 20 - > 10 - > 30 - > 80 - > 60 - > 90