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
来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        int n = preorder.length;        int preOrderLeft = 0;        int preOrderRight = n - 1;        int inOrderLeft = 0;        int inOrderRight = n - 1;
        TreeNode ans = helper(preorder, preOrderLeft, preOrderRight, inorder, inOrderLeft, inOrderRight);        return ans;    }
    public TreeNode helper(int[] preOrder, int preOrderLeft, int preOrderRight, int[] inOrder, int inOrderLeft, int inOrderRight) {        if (preOrderLeft > preOrderRight) {            return null;        }                int rootVal = preOrder[preOrderLeft];        int inOrderRootIndex = 0;        for (int i = 0; i < inOrder.length; i++) {            if (rootVal == inOrder[i]) {                inOrderRootIndex = i;                break;            }        }
        int leftSubTreeLength = inOrderRootIndex - inOrderLeft;
        TreeNode root = new TreeNode(rootVal);
        root.left = helper(preOrder, preOrderLeft + 1, preOrderLeft + leftSubTreeLength, inOrder, inOrderLeft, inOrderRootIndex - 1);        root.right = helper(preOrder, preOrderLeft + leftSubTreeLength + 1, preOrderRight, inOrder, inOrderRootIndex + 1, inOrderRight);
        return root;    }}