# 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 <= 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
## template
```java
public class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;
	TreeNode(int x) {
		val = x;
	}
}
public class Solution {
	public TreeNode buildTree(int[] preorder, int[] inorder) {
		if (preorder.length != inorder.length)
			return null;
		if (preorder.length == 0)
			return null;
		if (preorder.length == 1)
			return new TreeNode(preorder[0]);
		return buildTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
	}
	private TreeNode buildTree(int[] preorder, int[] inorder, int prei, int prej, int ini, int inj) {
		if (prei > prej || ini > inj || prei < 0 || prej >= preorder.length || ini < 0 || inj >= inorder.length)
			return null;
		if (prej - prei < 0)
			return null;
		if (prei == prej)
			return new TreeNode(preorder[prei]);
		TreeNode root = new TreeNode(preorder[prei]);
		int inFlag = 0;
		for (int i = ini; i <= inj; i++) {
			if (inorder[i] == root.val) {
				inFlag = i;
				break;
			}
		}
		int num_left = inFlag - ini;
		int num_right = inj - inFlag;
		root.left = buildTree(preorder, inorder, prei + 1, prei + num_left, ini, inFlag - 1);
		root.right = buildTree(preorder, inorder, prej - num_right + 1, prej, inFlag + 1, inj);
		return root;
	}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```