# 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
```