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build_binary_tree_problem
54 lines
1.4 KiB
Python
54 lines
1.4 KiB
Python
"""
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File: build_tree.py
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Created Time: 2023-07-15
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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 dfs(
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preorder: list[int],
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inorder: list[int],
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hmap: dict[int, int],
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i: int,
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l: int,
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r: int,
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) -> TreeNode | None:
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"""构建二叉树:分治"""
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# 子树区间为空时终止
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if r - l < 0:
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return None
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# 初始化根节点
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root = TreeNode(preorder[i])
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# 查询 m ,从而划分左右子树
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m = hmap[preorder[i]]
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# 子问题:构建左子树
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root.left = dfs(preorder, inorder, hmap, i + 1, l, m - 1)
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# 子问题:构建右子树
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root.right = dfs(preorder, inorder, hmap, i + 1 + m - l, m + 1, r)
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# 返回根节点
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return root
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def build_tree(preorder: list[int], inorder: list[int]) -> TreeNode | None:
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"""构建二叉树"""
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# 初始化哈希表,存储 inorder 元素到索引的映射
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hmap = {val: i for i, val in enumerate(inorder)}
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root = dfs(preorder, inorder, hmap, 0, 0, len(inorder) - 1)
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return root
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"""Driver Code"""
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if __name__ == "__main__":
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preorder = [3, 9, 2, 1, 7]
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inorder = [9, 3, 1, 2, 7]
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print(f"前序遍历 = {preorder}")
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print(f"中序遍历 = {inorder}")
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root = build_tree(preorder, inorder)
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print("构建的二叉树为:")
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print_tree(root)
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