perf: Optimize BTree search using binary search

This commit optimizes the search function in the BTree implementation
by replacing the linear search with a binary search algorithm. This
change significantly improves the search performance, especially for
large trees.

Implementation details:
- Modified the `search` method in the `BTree` struct
- Replaced the while loop with `binary_search` method on the `keys` vector

Complexity analysis:
- Previous implementation: O(n) per node, where n is the number of keys
- New implementation: O(log n) per node

Benchmark results:
- Environment: MacOS 14.5, Apple M1 Pro, 32 GB RAM
- Dataset: 1,000,000 random integers for insertion
- Search: 1,000,000 searches for the key 500,000
- Before:
  - Insertion: 3.002587333s
  - Search: 2.334683584s
- After:
  - Insertion: 2.998482583s
  - Search: 288.659458ms

Note: Insertion time remains largely unchanged, as expected. All
existing tests pass with the new implementation.

Benchmark code is not included in this commit.

Author: ie-Yoshisaur

src/data_structures/b_tree.rs

@@ -146,20 +146,16 @@ where
 
     pub fn search(&self, key: T) -> bool {
         let mut current_node = &self.root;
-        let mut index: isize;
         loop {
-            index = isize::try_from(current_node.keys.len()).ok().unwrap() - 1;
-            while index >= 0 && current_node.keys[index as usize] > key {
-                index -= 1;
-            }
-
-            let u_index: usize = usize::try_from(index + 1).ok().unwrap();
-            if index >= 0 && current_node.keys[u_index - 1] == key {
-                break true;
-            } else if current_node.is_leaf() {
-                break false;
-            } else {
-                current_node = &current_node.children[u_index];
+            match current_node.keys.binary_search(&key) {
+                Ok(_) => return true,
+                Err(index) => {
+                    if current_node.is_leaf() {
+                        return false;
+                    } else {
+                        current_node = &current_node.children[index];
+                    }
+                }
             }
         }
     }