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77 lines
2.5 KiB
77 lines
2.5 KiB
/*
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* File: preorder_traversal_iii_template.rs
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* Created Time: 2023-07-15
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* Author: sjinzh (sjinzh@gmail.com)
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*/
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include!("../include/include.rs");
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use std::{cell::RefCell, rc::Rc};
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use tree_node::{vec_to_tree, TreeNode};
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/* 判断当前状态是否为解 */
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fn is_solution(state: &mut Vec<Rc<RefCell<TreeNode>>>) -> bool {
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return !state.is_empty() && state.get(state.len() - 1).unwrap().borrow().val == 7;
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}
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/* 记录解 */
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fn record_solution(state: &mut Vec<Rc<RefCell<TreeNode>>>, res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>) {
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res.push(state.clone());
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}
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/* 判断在当前状态下,该选择是否合法 */
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fn is_valid(_: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNode>>) -> bool {
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return choice.borrow().val != 3;
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}
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/* 更新状态 */
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fn make_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNode>>) {
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state.push(choice);
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}
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/* 恢复状态 */
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fn undo_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, _: Rc<RefCell<TreeNode>>) {
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state.remove(state.len() - 1);
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}
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/* 回溯算法:例题三 */
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fn backtrack(state: &mut Vec<Rc<RefCell<TreeNode>>>, choices: &mut Vec<Rc<RefCell<TreeNode>>>, res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>) {
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// 检查是否为解
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if is_solution(state) {
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// 记录解
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record_solution(state, res);
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}
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// 遍历所有选择
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for choice in choices {
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// 剪枝:检查选择是否合法
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if is_valid(state, choice.clone()) {
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// 尝试:做出选择,更新状态
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make_choice(state, choice.clone());
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// 进行下一轮选择
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backtrack(state, &mut vec![choice.borrow().left.clone().unwrap(), choice.borrow().right.clone().unwrap()], res);
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// 回退:撤销选择,恢复到之前的状态
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undo_choice(state, choice.clone());
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}
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}
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}
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/* Driver Code */
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pub fn main() {
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let root = vec_to_tree([1, 7, 3, 4, 5, 6, 7].map(|x| Some(x)).to_vec());
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println!("初始化二叉树");
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print_util::print_tree(root.as_ref().unwrap());
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// 回溯算法
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let mut res = Vec::new();
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backtrack(&mut Vec::new(), &mut vec![root.unwrap()], &mut res);
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println!("\n输出所有根节点到节点 7 的路径,要求路径中不包含值为 3 的节点");
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for path in res {
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let mut vals = Vec::new();
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for node in path {
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vals.push(node.borrow().val)
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}
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println!("{:?}", vals);
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}
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}
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