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120 lines
3.3 KiB
120 lines
3.3 KiB
"""
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File: array_binary_tree.py
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Created Time: 2023-07-19
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Author: krahets (krahets@163.com)
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"""
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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from modules import TreeNode, list_to_tree, print_tree
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class ArrayBinaryTree:
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"""数组表示下的二叉树类"""
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def __init__(self, arr: list[int | None]):
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"""构造方法"""
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self._tree = list(arr)
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def size(self):
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"""列表容量"""
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return len(self._tree)
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def val(self, i: int) -> int | None:
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"""获取索引为 i 节点的值"""
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# 若索引越界，则返回 None ，代表空位
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if i < 0 or i >= self.size():
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return None
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return self._tree[i]
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def left(self, i: int) -> int | None:
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"""获取索引为 i 节点的左子节点的索引"""
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return 2 * i + 1
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def right(self, i: int) -> int | None:
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"""获取索引为 i 节点的右子节点的索引"""
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return 2 * i + 2
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def parent(self, i: int) -> int | None:
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"""获取索引为 i 节点的父节点的索引"""
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return (i - 1) // 2
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def level_order(self) -> list[int]:
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"""层序遍历"""
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self.res = []
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# 直接遍历数组
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for i in range(self.size()):
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if self.val(i) is not None:
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self.res.append(self.val(i))
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return self.res
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def dfs(self, i: int, order: str):
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"""深度优先遍历"""
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if self.val(i) is None:
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return
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# 前序遍历
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if order == "pre":
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self.res.append(self.val(i))
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self.dfs(self.left(i), order)
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# 中序遍历
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if order == "in":
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self.res.append(self.val(i))
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self.dfs(self.right(i), order)
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# 后序遍历
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if order == "post":
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self.res.append(self.val(i))
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def pre_order(self) -> list[int]:
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"""前序遍历"""
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self.res = []
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self.dfs(0, order="pre")
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return self.res
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def in_order(self) -> list[int]:
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"""中序遍历"""
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self.res = []
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self.dfs(0, order="in")
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return self.res
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def post_order(self) -> list[int]:
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"""后序遍历"""
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self.res = []
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self.dfs(0, order="post")
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return self.res
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"""Driver Code"""
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if __name__ == "__main__":
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# 初始化二叉树
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# 这里借助了一个从数组直接生成二叉树的函数
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arr = [1, 2, 3, 4, None, 6, 7, 8, 9, None, None, 12, None, None, 15]
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root = list_to_tree(arr)
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print("\n初始化二叉树\n")
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print("二叉树的数组表示：")
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print(arr)
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print("二叉树的链表表示：")
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print_tree(root)
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# 数组表示下的二叉树类
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abt = ArrayBinaryTree(arr)
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# 访问节点
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i = 1
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l, r, p = abt.left(i), abt.right(i), abt.parent(i)
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print(f"\n当前节点的索引为 {i} ，值为 {abt.val(i)}")
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print(f"其左子节点的索引为 {l} ，值为 {abt.val(l)}")
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print(f"其右子节点的索引为 {r} ，值为 {abt.val(r)}")
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print(f"其父节点的索引为 {p} ，值为 {abt.val(p)}")
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# 遍历树
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res = abt.level_order()
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print("\n层序遍历为：", res)
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res = abt.pre_order()
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print("前序遍历为：", res)
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res = abt.in_order()
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print("中序遍历为：", res)
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res = abt.post_order()
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print("后序遍历为：", res)
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