leetcode——从前序与中序遍历序列构造二叉树（java）

不是很理解这道题的解法，但还是对着题解敲了一遍，不会我们就把它背下来，没必要钻牛角尖。

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

/**
 * 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;
        if (n == 0) {
            return null;
        }
        int leftSize = indexOf(inorder, preorder[0]);
        int[] pre1 = Arrays.copyOfRange(preorder, 1, 1 + leftSize);
        int[] pre2 = Arrays.copyOfRange(preorder, 1 + leftSize, n);
        int[] in1 = Arrays.copyOfRange(inorder, 0, leftSize);
        int[] in2 = Arrays.copyOfRange(inorder, 1 + leftSize, n);
        TreeNode left = buildTree(pre1, in1);
        TreeNode right = buildTree(pre2, in2);
        return new TreeNode(preorder[0], left, right);
    }
    private int indexOf(int[] a, int x) {
        for (int i = 0; ; i++) {
            if (a[i] == x) {
                return i;
            }
        }
    }
}