在给定后序和中序遍历的情况下构建二叉树

Question

在this 链接之后，我修改了代码以在给定后序和中序遍历的情况下构建二叉树。但输出似乎产生了一些垃圾值。我不明白我哪里出错了。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 

Answer 1

node<int> * consTree(int * post, int * in, int postS, int postE, int inS, int inE){
    if(postS > postE)
    return NULL;

    int rootdata = post[postE];
    int indexI = -1;
    for(int i = inS; i <= inE; i++){
        if(rootdata == in[i]){
            indexI = i;
            break;
        }
    }
    
    int lpostS = postS;
    int linS = inS;
    int linE = indexI - 1;
    int lpostE = postS + indexI - inS - 1;
    int rpostS = postS + indexI - inS;
    int rpostE = postE - 1;
    int rinS = indexI + 1;
    int rinE = inE;

    node<int> * root = new node<int>(rootdata);
    root -> left = consTree(post,in,lpostS,lpostE,linS,linE);
    root -> right = consTree(post,in,rpostS,rpostE,rinS,rinE);
    return root;
}

基本上你需要考虑，当在递归调用中传递了正确的遍历顺序时，传递的数组保持不变，只是索引改变了..在这种情况下，假设我已经通过正确的顺序遍历（arr，6， 9）所以应该只在索引之间进行更改..而不是从 0（数组的起始索引）