Finding In-Order Successor in Binary Search tree Using Recursion

Question

I need to find the in-order successor in a binary search tree. For example, given a tree with format:

  4  
 / \  
2   7

and passing a search value of 2, the in-order successor would be 4.

Relevant Code:

template <typename T, typename Compare=std::less<T>>class BinarySearchTree {  
    private:
    struct Node {

        // Default constructor - does nothing
        Node() {}

        // Custom constructor provided for convenience
        Node(const T &datum_in, Node *left_in, Node *right_in)
        : datum(datum_in), left(left_in), right(right_in) { }

        T datum;
        Node *left;
        Node *right;
    };

And here's my function searching for the in-order successor

static Node * min_greater_than_impl(Node *node, const T &val, Compare less) {
    // base case: empty tree
    if (node == nullptr) {
        return nullptr;
    }
    // base case: no such element exists
    if (less((max_element_impl(node))->datum, val) || (!less((max_element_impl(node))->datum, val)
                                                  && !less(val, (max_element_impl(node))->datum))){
        return nullptr;
    }
    // recursive case: go right
    if (less(node->datum, val)) {
        if (node->right != nullptr) {
            return min_greater_than_impl(node->right, val, less);
        }
    }
    // recursive case: go left
    else if (less(val, node->datum)) {
        if (node->left != nullptr) {
            return min_greater_than_impl(node->left, val, less);
        }
    }
    // recurisve case: nequal to node, go right
    else {
        if (node->right != nullptr) {
            return min_greater_than_impl(node->right, val, less);
        }
    }
    return node;
}

Now, I believe the issue is that my function is not recognizing when it is actually at the desired node. I'm thinking that I need to check for this in maybe an extra base case or perhaps another else if statement at the end of the function. However, I can't think of how to do this without making a recursive call on the current node, but that would just end up in infinite loop.

Possible duplicate of [In Order Successor in Binary Search Tree](https://stackoverflow.com/questions/5471731/in-order-successor-in-binary-search-tree) — Bernhard Barker, Dec 31 '17 at 16:52

score 0 · Answer 1 · answered Dec 10 '16 at 09:22

The easiest way to do it is to add a parent pointer. Then to get "next", go up until you find the sibling.

If you can't use a parent, often for good reasons (e.g. the tree allows duplicate sub-branches), you have to keep your own stack of parent objects as you descend. To get the next object, if we are a left leaf, it is the parent, if we are a right leaf then if parent is a left leaf, it is the grandparent, otherwise we need to go up and check if grandparent is a left leaf of the great-grandparent. (If we hit the root, then our node was the last right leaf).

Finding In-Order Successor in Binary Search tree Using Recursion

1 Answers1