Binary Search Tree Iterator Sequel - Facebook Top Interview Questions


Problem Statement :


Implement a binary search tree iterator with the following methods:

next returns the next smallest element in the tree

hasnext returns whether there is a next element in the iterator

prev returns the next bigger element in the tree

hasprev returns whether there is a previous element in the iterator

For example, given the following tree root

   4
  / \
 2   7
    /
   5

Then we have

it = BSTIterator(root)

it.next() == 2

it.next() == 4

it.hasnext() == True

it.next() == 5

it.next() == 7

it.hasnext() == False

it.hasprev() == True

it.prev() == 5

Example 1

Input

methods = ["constructor", "hasnext", "hasnext", "hasprev", "hasprev", "next", "hasnext", "hasnext", 
"hasprev"]

arguments = [[[0, null, [2, [1, null, null], null]]], [], [], [], [], [], [], [], []]`

Output

[None, True, True, False, False, 0, True, True, False]


Solution :



title-img 




                        Solution in C++ :

class BSTIterator {  // Time: O(1) amortized, in worst case O(H), Space: O(N)
    private:
    stack<Tree*> next_stack;
    vector<int> history;
    int index;

    public:
    BSTIterator(Tree* root) {
        index = -1;

        while (root) {
            next_stack.push(root);
            root = root->left;
        }
    }

    int next() {
        if (index + 1 < history.size()) {
            index++;  // Move to next element in history and return it
            return history[index];
        }

        Tree* next_node = next_stack.top();
        next_stack.pop();
        int value = next_node->val;

        next_node = next_node->right;
        while (next_node) {
            next_stack.push(next_node);
            next_node = next_node->left;
        }

        index++;
        history.push_back(value);
        return value;
    }

    bool hasnext() {
        return !next_stack.empty();
    }

    int prev() {
        // Move to previous element in history and return it
        index--;
        return history[index];
    }

    bool hasprev() {
        return index > 0;
    }
};
                    




                        Solution in Python : 
                            
class BSTIterator:
    def __init__(self, root):
        self.nodes, self.cur = [], -1
        # get nodes
        def dfs(root):
            if root:  # inorder -----Left Root Right----
                dfs(root.left), self.nodes.append(root.val), dfs(root.right)

        dfs(root)
        self.size = len(self.nodes)

    def next(self):
        self.cur += 1
        return self.nodes[self.cur] if self.cur < self.size else None

    def hasnext(self):
        return self.cur < self.size - 1

    def prev(self):
        self.cur -= 1
        return self.nodes[self.cur] if self.cur >= 0 else None

    def hasprev(self):
        return self.cur > 0


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