**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 :

` ````
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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