# 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```
```

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <