/* * File: array_binary_tree.rs * Created Time: 2023-07-25 * Author: night-cruise (2586447362@qq.com) */ use hello_algo_rust::include::{print_util, tree_node}; /* 数组表示下的二叉树类 */ struct ArrayBinaryTree { tree: Vec>, } impl ArrayBinaryTree { /* 构造方法 */ fn new(arr: Vec>) -> Self { Self { tree: arr } } /* 列表容量 */ fn size(&self) -> i32 { self.tree.len() as i32 } /* 获取索引为 i 节点的值 */ fn val(&self, i: i32) -> Option { // 若索引越界,则返回 None ,代表空位 if i < 0 || i >= self.size() { None } else { self.tree[i as usize] } } /* 获取索引为 i 节点的左子节点的索引 */ fn left(&self, i: i32) -> i32 { 2 * i + 1 } /* 获取索引为 i 节点的右子节点的索引 */ fn right(&self, i: i32) -> i32 { 2 * i + 2 } /* 获取索引为 i 节点的父节点的索引 */ fn parent(&self, i: i32) -> i32 { (i - 1) / 2 } /* 层序遍历 */ fn level_order(&self) -> Vec { self.tree.iter().filter_map(|&x| x).collect() } /* 深度优先遍历 */ fn dfs(&self, i: i32, order: &'static str, res: &mut Vec) { if self.val(i).is_none() { return; } let val = self.val(i).unwrap(); // 前序遍历 if order == "pre" { res.push(val); } self.dfs(self.left(i), order, res); // 中序遍历 if order == "in" { res.push(val); } self.dfs(self.right(i), order, res); // 后序遍历 if order == "post" { res.push(val); } } /* 前序遍历 */ fn pre_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "pre", &mut res); res } /* 中序遍历 */ fn in_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "in", &mut res); res } /* 后序遍历 */ fn post_order(&self) -> Vec { let mut res = vec![]; self.dfs(0, "post", &mut res); res } } /* Driver Code */ fn main() { // 初始化二叉树 // 这里借助了一个从数组直接生成二叉树的函数 let arr = vec![ Some(1), Some(2), Some(3), Some(4), None, Some(6), Some(7), Some(8), Some(9), None, None, Some(12), None, None, Some(15), ]; let root = tree_node::vec_to_tree(arr.clone()).unwrap(); println!("\n初始化二叉树\n"); println!("二叉树的数组表示:"); println!( "[{}]", arr.iter() .map(|&val| if let Some(val) = val { format!("{val}") } else { "null".to_string() }) .collect::>() .join(", ") ); println!("二叉树的链表表示:"); print_util::print_tree(&root); // 数组表示下的二叉树类 let abt = ArrayBinaryTree::new(arr); // 访问节点 let i = 1; let l = abt.left(i); let r = abt.right(i); let p = abt.parent(i); println!( "\n当前节点的索引为 {} ,值为 {}", i, if let Some(val) = abt.val(i) { format!("{val}") } else { "null".to_string() } ); println!( "其左子节点的索引为 {} ,值为 {}", l, if let Some(val) = abt.val(l) { format!("{val}") } else { "null".to_string() } ); println!( "其右子节点的索引为 {} ,值为 {}", r, if let Some(val) = abt.val(r) { format!("{val}") } else { "null".to_string() } ); println!( "其父节点的索引为 {} ,值为 {}", p, if let Some(val) = abt.val(p) { format!("{val}") } else { "null".to_string() } ); // 遍历树 let mut res = abt.level_order(); println!("\n层序遍历为:{:?}", res); res = abt.pre_order(); println!("前序遍历为:{:?}", res); res = abt.in_order(); println!("中序遍历为:{:?}", res); res = abt.post_order(); println!("后序遍历为:{:?}", res); }