Q：根据二叉树的前序和中序遍历结果重建二叉树
A：前序遍历的第一个节点是父节点，中序遍历中这个节点的左边即是该父节点的左子树，长度为lenLeft，右边就是该父节点的右子树，长度为lenRight。前序遍历中父节点右边lenLeft长度为左子树，再往右lenRight长度为右子树。
代码如下

    static class Node
    {
        public int data;
        public Node left;
        public Node right;
    }
    public static Node create(int[] preArr,int[] midArr)
    {
        if (preArr==null||midArr==null||preArr.length==0||midArr.length==0)
        {
            return null;
        }
        return createHelper(preArr,0,preArr.length-1,midArr,0,midArr.length-1);
    }
    public static Node 
        createHelper(int[] preArr,int preStart,int preEnd,int[] midArr,int midStart,int midEnd)
    {
        Node root=new Node();
        //根节点就是前序遍历中开头的节点
        root.data=preArr[preStart];
        if (preStart==preEnd)
        {
            if (midStart==midEnd&&preArr[preStart]==midArr[midStart])
            {
                return root;
            }
            else
            {
                throw new RuntimeException("input valid");
            }
        }
        //找到中序遍历中根节点的位置:temp
        int temp=midStart;
        while (temp<=midEnd&&midArr[temp]!=preArr[preStart])
        {
            temp++;
        }
        if (temp>midEnd)
        {
            throw new RuntimeException("input valid");
        }
        int leftLen=temp-midStart;
        if (leftLen>0)
        {
            root.left=createHelper(preArr,preStart+1,preStart+leftLen,midArr,midStart,temp-1);
        }
        if (leftLen<midEnd-midStart)
        {
            root.right=createHelper(preArr,preStart+leftLen+1,preEnd,midArr,temp+1,midEnd);
        }
            return root;
    }
    public static void main(String[] args) 
    {
        int[] preArr={1,2,4,7,3,5,6,8};
        int[] midArr={4,7,2,1,5,3,8,6};
        Node t=create(preArr,midArr);
        preOrder(t);
        System.out.println();
        midOrder(t);
        System.out.println();
        postOrder(t);
        System.out.println();
        doubleStackPostOrder(t);
    }