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70 lines
2.0 KiB
Python
70 lines
2.0 KiB
Python
"""
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File: tree_node.py
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Created Time: 2021-12-11
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Author: krahets (krahets@163.com)
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"""
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from collections import deque
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class TreeNode:
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"""二叉树节点类"""
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def __init__(self, val: int = 0):
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self.val: int = val # 节点值
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self.height: int = 0 # 节点高度
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self.left: TreeNode | None = None # 左子节点引用
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self.right: TreeNode | None = None # 右子节点引用
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# 序列化编码规则请参考：
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# https://www.hello-algo.com/chapter_tree/array_representation_of_tree/
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# 二叉树的数组表示：
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# [1, 2, 3, 4, None, 6, 7, 8, 9, None, None, 12, None, None, 15]
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# 二叉树的链表表示：
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# /——— 15
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# /——— 7
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# /——— 3
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# | \——— 6
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# | \——— 12
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# ——— 1
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# \——— 2
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# | /——— 9
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# \——— 4
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# \——— 8
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def list_to_tree_dfs(arr: list[int], i: int) -> TreeNode | None:
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"""将列表反序列化为二叉树：递归"""
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# 如果索引超出数组长度，或者对应的元素为 None ，则返回 None
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if i < 0 or i >= len(arr) or arr[i] is None:
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return None
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# 构建当前节点
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root = TreeNode(arr[i])
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# 递归构建左右子树
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root.left = list_to_tree_dfs(arr, 2 * i + 1)
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root.right = list_to_tree_dfs(arr, 2 * i + 2)
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return root
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def list_to_tree(arr: list[int]) -> TreeNode | None:
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"""将列表反序列化为二叉树"""
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return list_to_tree_dfs(arr, 0)
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def tree_to_list_dfs(root: TreeNode, i: int, res: list[int]) -> list[int]:
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"""将二叉树序列化为列表：递归"""
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if root is None:
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return
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if i >= len(res):
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res += [None] * (i - len(res) + 1)
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res[i] = root.val
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tree_to_list_dfs(root.left, 2 * i + 1, res)
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tree_to_list_dfs(root.right, 2 * i + 2, res)
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def tree_to_list(root: TreeNode | None) -> list[int]:
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"""将二叉树序列化为列表"""
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res = []
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tree_to_list_dfs(root, 0, res)
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return res
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