给出后序和顺序遍历的二叉树构建

Question

在this链接之后，我修改了代码以根据后序和顺序遍历构建二叉树。但输出似乎产生了一些垃圾值。我无法理解我哪里出错了。前序遍历在开头有根，后序遍历在结尾有根。使用该逻辑，我修改了代码。如下：

struct tree* buildTree2(char in[], char post[], int inStrt, int inEnd)
{
    static int postIndex = sizeof(post)/sizeof(post[0]) - 1; //postorder index
    if(inStrt > inEnd) 
        return NULL;
    struct tree* tNode = newNode(post[postIndex--]);
    if(inStrt == inEnd)
        return tNode;
    else
    {
        int inIndex = search(in, inStrt, inEnd, tNode->data);
        tNode->left = buildTree(in, post, inStrt, inIndex-1); 
        tNode->right = buildTree(in, post, inIndex+1, inEnd);
        return tNode;
    }
}

输出：

 The inorder traversal of the build tree (using preorder traversal) is : D B E A F C 
 The inorder traversal of the build tree (using postorder traversal) is : D B I R O  7 = R N T V _ G D X  t u o . a / . 

修改后的代码：

struct tree* buildTree2(char in[], char post[], int inStrt, int inEnd, int postIndex)
{
    //printf("\n %d ",postIndex);
    if(inStrt > inEnd) //termination condition for buildTree(in, post, inIndex+1, inEnd)
        return NULL;
    struct tree* tNode = newNode(post[postIndex--]);
    //check if node has children
    if(inStrt == inEnd)
        return tNode;
    else
    {
        //get the index of the postorder variable in the inorder traversal
        int inIndex = search(in, inStrt, inEnd, tNode->data);
        //Recursively build the tree
        tNode->left = buildTree2(in, post, inStrt, inIndex-1, postIndex); 
          //The inIndex value points to the tNode. So less than that is left sub tree and more than that is the right sub tree
        tNode->right = buildTree2(in, post, inIndex+1, inEnd, postIndex);
        return tNode;
    }
}

输出：

The inorder traversal of the build tree (using preorder traversal) is : D B E A F C 
 The inorder traversal of the build tree (using postorder traversal) is : E B E