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@ -11,7 +11,7 @@ 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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return !state.is_empty() && state.last().unwrap().borrow().val == 7;
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}
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/* 记录解 */
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@ -23,8 +23,8 @@ fn record_solution(
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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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fn is_valid(_: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Option<&Rc<RefCell<TreeNode>>>) -> bool {
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return choice.is_some() && choice.unwrap().borrow().val != 3;
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}
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/* 更新状态 */
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@ -34,13 +34,13 @@ fn make_choice(state: &mut Vec<Rc<RefCell<TreeNode>>>, choice: Rc<RefCell<TreeNo
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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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state.pop();
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}
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/* 回溯算法:例题三 */
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fn backtrack(
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state: &mut Vec<Rc<RefCell<TreeNode>>>,
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choices: &mut Vec<Rc<RefCell<TreeNode>>>,
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choices: &Vec<Option<&Rc<RefCell<TreeNode>>>>,
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res: &mut Vec<Vec<Rc<RefCell<TreeNode>>>>,
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) {
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// 检查是否为解
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@ -49,22 +49,22 @@ fn backtrack(
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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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for &choice in choices.iter() {
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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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if is_valid(state, choice) {
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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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make_choice(state, choice.unwrap().clone());
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// 进行下一轮选择
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backtrack(
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state,
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&mut vec![
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choice.borrow().left.clone().unwrap(),
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choice.borrow().right.clone().unwrap(),
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&vec![
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choice.unwrap().borrow().left.as_ref(),
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choice.unwrap().borrow().right.as_ref(),
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],
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res,
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);
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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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undo_choice(state, choice.unwrap().clone());
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}
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}
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}
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@ -77,7 +77,7 @@ pub fn main() {
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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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backtrack(&mut Vec::new(), &mut vec![root.as_ref()], &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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