Binary Tree Data Structure
Rust Programming Language
Problem
Rust program that implements a binary tree data structure.
Input
// Define a struct for a binary tree nodestruct Node {value: i32,left: Option<Box<Node>>,right: Option<Box<Node>>,}// Define a function to insert a new node into the binary treefn insert_node(root: &mut Option<Box<Node>>, value: i32) {if let Some(ref mut node) = root {if value < node.value {insert_node(&mut node.left, value);} else {insert_node(&mut node.right, value);}} else {*root = Some(Box::new(Node {value,left: None,right: None,}));}}// Define a function to print out the contents of the binary tree in orderfn inorder_traversal(root: &Option<Box<Node>>) {if let Some(ref node) = root {inorder_traversal(&node.left);println!("{}", node.value);inorder_traversal(&node.right);}}fn main() {// Create an empty root nodelet mut root = None;// Insert some values into the binary treeinsert_node(&mut root, 5);insert_node(&mut root, 3);insert_node(&mut root, 7);insert_node(&mut root, 1);insert_node(&mut root, 9);// Print out the contents of the binary tree in orderinorder_traversal(&root);}{codeBox}
Output
13579{codeBox}
Explanation
In this example, the Node struct represents a single node in the binary tree, containing a value and pointers to its left and right child nodes. The insert_node function recursively inserts new nodes into the binary tree based on their values, and the inorder_traversal function recursively prints out the contents of the binary tree in order.
The main function creates an empty root node, inserts some values into the binary tree, and then prints out the contents of the binary tree in order.
This is the inorder traversal of the binary tree that was created in the main function, which includes the values 1, 3, 5, 7, and 9. The inorder traversal visits the left subtree first, then the root node, and then the right subtree. In this case, the left subtree of the root node contains the values 1 and 3, the right subtree contains the values 7 and 9, and the root node itself contains the value 5. Therefore, the inorder traversal visits the nodes in the order 1, 3, 5, 7, 9.