从前序与中序遍历序列构造二叉树

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

递归解法：

/**
 * 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 build(preorder,inorder,0,preorder.length-1,0,inorder.length-1);
    }

    private TreeNode build(int[] preorder, int[] inorder,int preStart,int preEnd,int inStart,int inEnd){
        if (preStart > preEnd || inStart > inEnd || preStart >= preorder.length || inStart >= inorder.length)
            return null;
        int rootVal = preorder[preStart];
        int offset = 0;
        while(inorder[offset+inStart] != rootVal){
            offset++;
        }
        TreeNode root = new TreeNode(rootVal);
        root.left = build(preorder,inorder,preStart+1,preStart+offset,inStart,inStart+offset);
        root.right = build(preorder,inorder,preStart+offset+1,preEnd,inStart+offset+1,inEnd);
        return root;
    }
}