# Search a node in Binary search tree

Category: Data Structures

In this tutorial, we are going to learn that how to search a node in a Binary search tree using recursive and iterative solution.

While searching a node in a Binary Tree, we traverse each node one by one and check if required node is available or not.

But in case of Binary Search Tree, we are not required to traverse and check each and every node. Here we can compare the root node and if it is equal to the root node then search is complete otherwise check if node to be searched is less than or greater than the root node.

If the node to be searched is less than root node then ignore the right subtree and continue with the same approach in left subtree or if that node is greater than root node then ignore the left subtree and continue with the same approach in right subtree.

If the node to be searched is available in the BST then return `True` otherwise return `False`.

Example -

``````Input:
5
/   \
3     8
/ \   / \
1   4 6   9

Node to be searched: 6

Output: True
``````

We can solve this problem in two ways-

## Recursive solution

Algorithm

``````Find_Recur(root, data)
1.  if root is null then return False
2.  if root.data == data then return True
3.  else
4.    if data < root.data
5.      return Find_Recur(root.left, data)
6.    else
7.      return Find_Recur(root.right, data)
``````
Python
``````class Node:
def __init__(self, data):
self.left = None
self.data = data
self.right = None

def find(root, data):
if not root: return False
if root.data == data: return True
else:
if data < root.data:
return find(root.left, data)
else:
return find(root.right, data)

bst_root = Node(10)
bst_root.left = Node(6)
bst_root.right = Node(16)
bst_root.left.left = Node(4)
bst_root.left.right = Node(9)
bst_root.right.left = Node(13)
bst_root.right.right = Node(20)
print(find(bst_root, 16))
``````
JavaScript
``````class Node {
constructor(data) {
this.left = null;
this.data = data;
this.right = null;
}
}

function find(root, data) {
if (!root) return false;
if (root.data == data) return true;
else {
if (data < root.data) return find(root.left, data);
else return find(root.right, data);
}
}

const bstRoot = new Node(10);
bstRoot.left = new Node(6);
bstRoot.right = new Node(16);
bstRoot.left.left = new Node(4);
bstRoot.left.right = new Node(9);
bstRoot.right.left = new Node(13);
bstRoot.right.right = new Node(20);
console.log(find(bstRoot, 16));
``````

Output

``````True
``````

## Iterative solution

Algorithm

``````Find_Iter(root, data)
1.  node = root
2.  while node is not null
3.    if node.data == data then
4.      return True
5.    if data < node.data
6.      node = node.left
7.    else
8.      node = node.right
8.  return False
``````
Python
``````class Node:
def __init__(self, data):
self.left = None
self.data = data
self.right = None

def find(root, data):
node = root
while node:
if node.data == data: return True
if data < node.data:
node = node.left
else:
node = node.right
return False

bst_root = Node(10)
bst_root.left = Node(6)
bst_root.right = Node(16)
bst_root.left.left = Node(4)
bst_root.left.right = Node(9)
bst_root.right.left = Node(13)
bst_root.right.right = Node(20)
print(find(bst_root, 16))
``````
JavaScript
``````class Node {
constructor(data) {
this.left = null;
this.data = data;
this.right = null;
}
}

function find(root, data) {
let node = root;
while (node != null) {
if (node.data == data) return true;
if (data < node.data) node = node.left;
else node = node.right;
}
return false;
}

const bstRoot = new Node(10);
bstRoot.left = new Node(6);
bstRoot.right = new Node(16);
bstRoot.left.left = new Node(4);
bstRoot.left.right = new Node(9);
bstRoot.right.left = new Node(13);
bstRoot.right.right = new Node(20);
console.log(find(bstRoot, 16));
``````

Output

``````True
``````

Time complexity: The time complexity of both of these solution will be O(log n) if the tree is balanced but in worst case the time complete will be O(n) if the tree is not balanced i.e, tree becomes skewed.