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105.从前序与中序遍历序列构造二叉树 ​

https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

img
输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder 和 inorder 均 无重复 元素
  • inorder 均出现在 preorder
  • preorder 保证 为二叉树的前序遍历序列
  • inorder 保证 为二叉树的中序遍历序列
python
from typing import List, Optional
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if not preorder or not inorder:
            return None

        # The first element in preorder is the root
        root_val = preorder[0]
        root = TreeNode(root_val)

        # Find the index of the root in inorder
        root_index = inorder.index(root_val)

        # Recursively build the left and right subtrees
        root.left = self.buildTree(preorder[1:1 + root_index], inorder[:root_index])
        root.right = self.buildTree(preorder[1 + root_index:], inorder[root_index + 1:])

        return root

if __name__ == '__main__':
    solution = Solution()
    preorder = [3, 9, 20, 15, 7]
    inorder = [9, 3, 15, 20, 7]
    root = solution.buildTree(preorder, inorder)
    # The output tree is [3, 9, 20, None, None, 15, 7]