105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.Note:
You may assume that duplicates do not exist in the tree.
已知二叉树的先序遍历和中序遍历，还原二叉树。二叉树中无重复值，程序如下所示：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return recover(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
}

public TreeNode recover(int[] preorder, int preBegin, int preEnd,
int[] inorder, int inBegin, int inEnd){
if (preBegin > preEnd||inBegin > inEnd){
return null;
}
TreeNode root = new TreeNode(preorder[preBegin]);
int i = 0;
for (i = 0; i <= inEnd; ++ i){
if (inorder[i] == preorder[preBegin]){
break;
}
}
root.left = recover(preorder, preBegin + 1, preBegin + (i - inBegin), inorder, inBegin, i - 1);
root.right = recover(preorder, preBegin + (i - inBegin ) + 1, preEnd, inorder, i + 1, inEnd);
return root;
}
}