尽管有真实值，C中的函数始终返回0

Question

typedef struct BinaryTreeNode {
int data;
BinaryTreeNode * left;
BinaryTreeNode * right;
} BinaryTreeNode;

int isElementInBinaryTree(BinaryTreeNode *root, int search_item) {
    if(root) {
        if(search_item == root -> data) return 1;
        isElementInBinaryTree(root -> left, search_item);
        isElementInBinaryTree(root -> right, search_item);
    }
}

int main() {
    BinaryTreeNode one = {1, NULL, NULL}; // root of the binary tree
    BinaryTreeNode two = {2, NULL, NULL};
    BinaryTreeNode three = {3, NULL, NULL};
    BinaryTreeNode four = {4, NULL, NULL};
    BinaryTreeNode five = {5, NULL, NULL};
    BinaryTreeNode six = {6, NULL, NULL};
    BinaryTreeNode seven = {7, NULL, NULL};

    one.left = &two;
    one.right = &three;

    two.left = &four;
    two.right = &five;

    three.left = &six;
    three.right = &seven;

    printf("%d ", isElementInBinaryTree(&one, 4));
    printf("\n");

    return 0;
}

I am writing a function called isElementInBinaryTree that returns 1 (true) is the element exists and 0 otherwise. I don't understand why the function is always returning 0 despite the fact that the number 4 exists in the binary tree ?

Answer 1

Your code doesn't actually return anything when it recurses, so the 'return value' is typically whatever is left over in the register used for that purpose.

This code

if(root) {
        if(search_item == root -> data) return 1;
        isElementInBinaryTree(root -> left, search_item);
        isElementInBinaryTree(root -> right, search_item);
    }

needs to look more like this

if(root) 
{
    if(search_item == root -> data) return 1;

    if (isElementInBinaryTree(root -> left, search_item)) return 1;
    return isElementInBinaryTree(root -> right, search_item);
}
return 0;

The final return 0; ensures you return something sensible when a NULL pointer is provided.

When you compile your code it should be displaying warnings about a typed function terminating without a return statement. You need to pay attention to those warnings and resolve them all.

The whole thing can actually be reduced to a single (though less clear) line of code

return root && ((search_item == root->data) ||
                isElementInBinaryTree(root->left, search_item) ||
                isElementInBinaryTree(root->right, search_item));

which relies on shortcut evaluation to only go so far as needed.

Answer 2

You're not returning anything when you perform the recursive calls. So it will only return 1 if the item is found at the root of the tree.

Also, you don't return anything when you reach the bottom of the tree without finding anyhthing, this needs to return 0 to indicate failure.

int isElementInBinaryTree(BinaryTreeNode *root, int search_item) {
    if(root) {
        if(search_item == root -> data) {
            return 1;
        }
        return isElementInBinaryTree(root -> left, search_item) || 
               isElementInBinaryTree(root -> right, search_item);
    } else {
        return 0;
    }
}