Question

我正在尝试在Rust中实现一个简单的二叉搜索树，但我很难确定插入节点的问题。我正在使用以下数据结构和功能。

enum BinaryTree<T> {
    Leaf(T),
    Branch(T, Box<BinaryTree<T>>, Box<BinaryTree<T>>),
    Null,
}

fn createBinarySearchTree(vector: Vec<int>) -> BinaryTree<int> {        
    fn insertNode(val: int, btree: &BinaryTree<int>) -> BinaryTree<int> {
        match btree {
            &Leaf(tval) if val > tval => Branch(tval, box Null, box Leaf(val)),     
            &Leaf(tval) if val < tval => Branch(tval, box Leaf(val), box Null),
            &Branch(tval, box ref left, box ref right) if val > tval => insertNode(val,right),
            &Branch(tval, box ref left, box ref right) if val < tval => insertNode(val,left),
            &Null => Leaf(val),
            &Leaf(lval) if val == lval => Leaf(val),
            &Branch(lval, box ref left, box ref right) if val == lval => fail!("already has a node with {}", lval),
            _ => Null,
        }
    }

    let mut tree = Null;
    for v in vector.iter() {
        tree = insertNode(*v, &tree);
    }

    let immuTree = tree;
    immuTree
}

fn printTree(tree: &BinaryTree<int>) {
    fn innerPrint(prefix: &str, tree: &BinaryTree<int>, level: int) {
        let lvDesc = format!("lv {}", level);
        match tree {
            &Leaf(val) => println!("{}-{} leaf: {}", lvDesc, prefix, val),
            &Branch(val, box ref left, box ref right) => {
                println!("{}-{} node: {}", lvDesc, prefix, val);
                innerPrint("left branch <-", left, level + 1);
                innerPrint("right branch ->", right, level + 1);
            },
            &Null => println!("end"),
        }
    }
    innerPrint("root", tree, 0);
}

在调用printTree(&createBinarySearchTree(vec![43,2,45,7,72,28,34,33]))时，树只打印出33,34，不幸的是我无法调试，因为使用调试信息进行编译会导致编译器错误。此外，当我在插入时在分支上匹配时，我试图返回一个分支，但这需要我克隆叶子/给予所有权的方式，我只是无法绕过我的脑袋。所以任何帮助都会非常感激

干杯

Answer 1

我相信这些分支是错的：

&Branch(tval, box ref left, box ref right) if val > tval => insertNode(val, right),
&Branch(tval, box ref left, box ref right) if val < tval => insertNode(val, left),

由于您正在改变原始树，因此每个分支都会丢失原始树根。假设修复（未经测试）：

&Branch(tval, box ref left, box ref right) if val > tval => Branch(tval, left, insertNode(val, right)),
&Branch(tval, box ref left, box ref right) if val < tval => Branch(tval, insertNode(val, left), right),

修改

这个想法是正确的，但是你是正确的，Rust抱怨要离开＆amp;一个模式守卫背后的指针，所以我不得不在里面做另一场比赛（结果更好）。我也无法忽略命名，所以我按照Rust编码风格清理它：

use std::fmt::Show; enum BinaryTree<T> { Leaf(T), Branch(T, Box<BinaryTree<T>>, Box<BinaryTree<T>>), Null, } fn create_binary_search_tree(vector: Vec<int>) -> BinaryTree<int> { fn insert_node<T: Copy + Ord + Show>(val: T, btree: BinaryTree<T>) -> BinaryTree<T> { match btree { Leaf(tval) if val > tval => Branch(tval, box Null, box Leaf(val)), Leaf(tval) if val < tval => Branch(tval, box Leaf(val), box Null), Branch(tval, left, right) => match val.cmp(&tval) { Greater => Branch(tval, left, box insert_node(val, *right)), Less => Branch(tval, box insert_node(val, *left), right), Equal => fail!("already has a node with {}", tval), }, Null => Leaf(val), Leaf(lval) if val == lval => Leaf(val), _ => Null, } } let mut tree = Null; for v in vector.iter() { tree = insert_node(*v, tree); } let immuTree = tree; immuTree } fn print_tree(tree: &BinaryTree<int>) { fn inner_print(prefix: &str, tree: &BinaryTree<int>, level: int) { let lvDesc = format!("lv {}", level); match tree { &Leaf(val) => println!("{}-{} leaf: {}", lvDesc, prefix, val), &Branch(val, box ref left, box ref right) => { println!("{}-{} node: {}", lvDesc, prefix, val); inner_print("left branch <-", left, level + 1); inner_print("right branch ->", right, level + 1); }, &Null => println!("end"), } } inner_print("root", tree, 0); } fn main() { print_tree(&create_binary_search_tree(vec![43, 2, 45, 7, 72, 28, 34, 33])); }

我验证此代码适用于“http://play.rust-lang.org/”

生锈二叉树实现的困难

1 个答案: