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68 lines
1.8 KiB
Python
68 lines
1.8 KiB
Python
"""
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File: backtrack_find_constrained_path.py
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Created Time: 2023-04-15
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Author: Krahets (krahets@163.com)
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"""
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from modules import *
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def is_solution(state: list[TreeNode]) -> bool:
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"""判断当前状态是否为解"""
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return state and state[-1].val == 7
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def record_solution(state: list[TreeNode], res: list[list[TreeNode]]):
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"""记录解"""
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res.append(list(state))
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def is_valid(state: list[TreeNode], choice: TreeNode) -> bool:
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"""判断在当前状态下，该选择是否合法"""
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return choice is not None and choice.val != 3
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def make_choice(state: list[TreeNode], choice: TreeNode):
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"""更新状态"""
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state.append(choice)
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def undo_choice(state: list[TreeNode], choice: TreeNode):
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"""恢复状态"""
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state.pop()
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def backtrack(state: list[TreeNode], choices: list[TreeNode], res: list[list[TreeNode]]):
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"""回溯算法"""
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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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return
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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):
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# 尝试：做出选择，更新状态
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make_choice(state, choice)
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backtrack(state, [choice.left, choice.right], 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)
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"""Driver Code"""
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if __name__ == "__main__":
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root = list_to_tree([1, 7, 3, 4, 5, 6, 7])
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print("\n初始化二叉树")
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print_tree(root)
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# 回溯算法
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res = []
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backtrack(state=[], choices=[root], res=res)
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print("\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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print([node.val for node in path])
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