已知二叉树前序和中序，算法写出后续遍历的结果

package com.S;

import java.util.Arrays;

public class Tree {

    int data;
    Tree left;
    Tree right;

    public Tree(int data){
        this.data = data;
    }
    public Tree(int data, Tree left, Tree right){
        this.data = data;
        this.left = left;
        this.right = right;
    }

    public void afterShow(){
        //左
        if(left != null){
            left.afterShow();
        }
        //右
        if(right != null){
            right.afterShow();
        }
        //自己
        System.out.println(data);
    }

    /**
     * 重建树
     * @param preOrders  树的前序
     * @param inOrders   树的中序
     * @return
     */
    public Tree buildTree(int[] preOrders, int[] inOrders){
        //递归截止条件，有一个为空，就说明没有子节点了
        if(preOrders.length == 0 || inOrders.length == 0){
            return null;
        }
        //每轮的根节点就是preOrders的第一个
        Tree root = new Tree(preOrders[0]);
        //根据inOrders查找到根节点的下标，返回index
        int index = search(inOrders, root.data);
        if(index >= 0){
            //Arrays.copyOfRange(preOrders, 1, index+1)，截取preOrders，从1到index+1，前包后不包
            root.left = buildTree(Arrays.copyOfRange(preOrders, 1, index+1), Arrays.copyOfRange(inOrders, 0, index));
            root.right = buildTree(Arrays.copyOfRange(preOrders, index+1, preOrders.length), Arrays.copyOfRange(inOrders, index+1, inOrders.length));
        }
        return root;
    }

    /**
     * 根据inOrders查找根节点的下标，目的是找到左子树和右子树的分界
     * @param inOrders  中序
     * @param data   根节点
     * @return
     */
    public int search(int[] inOrders, int data){
        for (int i = 0; i < inOrders.length; i++) {
            if(data == inOrders[i]){
                return i;
            }
        }
        return -1;
    }


    public static void main(String[] args) {
       int[] preOrders = {1,2,4,5,3,6,7};
       int[] inOrders = {4,2,5,1,6,3,7};
       new Tree(preOrders[0]).buildTree(preOrders, inOrders).afterShow();
    }

}