# 从前序与中序遍历序列构造二叉树
给定一棵树的前序遍历 preorder 与中序遍历 inorder。请构造二叉树并返回其根节点。
示例 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
提示:
1 <= preorder.length <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorder 和 inorder 均无重复元素inorder 均出现在 preorderpreorder 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
## template
```python
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def buildTree(self, preorder, inorder):
if not preorder:
return None
root = TreeNode(preorder[0])
i = inorder.index(root.val)
root.left = self.buildTree(preorder[1 : i + 1], inorder[:i])
root.right = self.buildTree(preorder[i + 1 :], inorder[i + 1 :])
return root
```
## 答案
```python
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## 选项
### A
```python
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### B
```python
```
### C
```python
```