# 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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