根据一棵树的中序遍历与后序遍历构造二叉树。

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

//    3
//   / \
//  9  20
//    /  \
//   15   7

    private int index2;
    public TreeNode buildTree(int[] preorder, int[] postorder) {
        index2 = 0;
        //先逆置
        int length = postorder.length;
        for (int i = 0; i < length / 2; i++) {
            int temp = postorder[i];
            postorder[i] = postorder[length - 1 - i];
            postorder[length - 1 - i] = temp;
        }
        return buildTreeHelper(preorder,postorder,0,postorder.length);
    }

    private TreeNode buildTreeHelper(int[] preorder, int[] postorder, int left, int right) {
        if (left >= right){
            //中序遍历结果为空，这个数就是空树
            return null;
        }
        if (index2 >= preorder.length){
            return null;
        }
        TreeNode root = new TreeNode((char) preorder[index2]);
        index2++;
        int pos = find(postorder,left,right,root.val);
        root.right = buildTreeHelper(preorder,postorder,pos+1,right);
        root.left = buildTreeHelper(preorder,postorder,left,pos);

        //一个树的先序遍历的镜像和后序遍历的逆置相同，，，，根右左
        //所以先逆置后序遍历，再调整左右根的打印位置

        return root;
    }

    private int find(int[] inorder, int left, int right, int toFind) {
        for (int i = left; i < right; i++){
            if (inorder[i] == toFind){
                return i;
            }
        }
        return -1;
    }