C++ Reference - SomeType* &val vs. SomeType* val

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I'm solving LeetCode 783. Minimum Distance Between BST Nodes and I've noticed that the difference between a correct solution and an incorrect solution is a reference (&) at my function call, as follows:

Correct Solution:

class Solution {
public:
    void traverse(TreeNode* root, TreeNode* &curr, int &sol){
        if (root == nullptr) return;
        traverse(root->left, curr, sol);        
        if(curr) sol = min(sol, abs(root->val - curr->val));
        curr = root;
        traverse(root->right, curr, sol);
    }

    int minDiffInBST(TreeNode* root) {
        int sol = INT_MAX;
        TreeNode* curr = nullptr;
        traverse(root, curr, sol);
        return sol;
    }
};

Incorrect Solution:

class Solution {
public:
    void traverse(TreeNode* root, TreeNode* curr, int &sol){
        //Exactly the same as above!
};

As a student, this is the first time I have encountered this case related to Pointers and References. I'll appreciate any explanation of this difference.

pointers pass-by-reference c++
2 Answers

The answer is that without reference (void traverse(TreeNode* root, TreeNode* curr, int &sol){...}) curr value will not be updated for the future calls of the function (which will be executed from the call-stack).

But when there is a reference (void traverse(TreeNode* root, TreeNode* &curr, int &sol){...}) curr value will be updated and will be used for the next calls until termination of the program.

user_89927100

If you do this

void foo(int * inner_ptr) {
   ptr++;
}

int main() {
   int arr[5] = {1, 2, 3, 4, 5};
   int outer_ptr = &arr[1];
   foo(outer_ptr);
}

the outer_ptr will still be equal to &arr[1].

You only changed the inner_ptr, the copy of the outer_ptr.

You can change the thing it points to.

void foo(int * inner_ptr) {
   (*ptr) = 42;
}

But not the outer_ptr itself

Therefore you need either this signature: (with reference)

void foo(int * & inner_ptr);

or this signature: (pointer to pointer) (in this case you would work with it differently in the function body tho)

void foo(int * * inner_ptr);
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