# Insert into a Binary Search Tree – Leetcode Solution

In this post, we are going to solve the 701. Insert into a Binary Search Tree problem of Leetcode. This problem 701. Insert into a Binary Search Tree is a Leetcode medium level problem. Let’s see the code, 701. Insert into a Binary Search Tree – Leetcode Solution.

## Problem

You are given the `root` node of a binary search tree (BST) and a `value` to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

### Example 1 :

``````Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]``````

### Example 2 :

``````Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]``````

### Example 3 :

``````Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]``````

### Constraints

• The number of nodes in the tree will be in the range `[0, 104]`.
• `-108 <= Node.val <= 108`
• All the values `Node.val` are unique.
• `-108 <= val <= 108`
• It’s guaranteed that `val` does not exist in the original BST.

Now, let’s see the code of 701. Insert into a Binary Search Tree – Leetcode Solution.

# Insert into a Binary Search Tree– Leetcode Solution

### 701. Insert into a Binary Search Tree – Solution in Java

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if(root == null) return new TreeNode(val);
TreeNode curr = root;
while(true){
if(curr.val >= val){
if(curr.left != null) curr = curr.left;
else {
curr.left = new TreeNode(val);
break;
}
}
else{
if(curr.right != null) curr = curr.right;
else {
curr.right = new TreeNode(val);
break;
}
}

}
return root;
}
}```

### 701. Insert into a Binary Search Tree – Solution in C++

```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* insertIntoBST(TreeNode* root, int val) {
if(!root) return new TreeNode(val);

auto curr = root;

while(true){
if(curr->val < val){
if(curr->right) curr = curr->right;
else {
curr->right = new TreeNode(val);
break;
}
}
else{
if(curr->left) curr = curr->left;
else{
curr->left = new TreeNode(val);
break;
}
}
}
return root;
}
};```

### 701. Insert into a Binary Search Tree– Solution in Python

```# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root:
return TreeNode(val)
tempRoot = root
while root:
if root.val < val:
if not root.right:
root.right = TreeNode(val)
break
root = root.right
else:
if not root.left:
root.left = TreeNode(val)
break
root = root.left
return tempRoot```

Note: This problem 701. Insert into a Binary Search Tree is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.