# 701. Insert into a Binary Search Tree

## Description

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]
Explanation: Another accepted tree is:



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.

## Solutions

DFS.

• /**
* 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);
}
if (root.val < val) {
root.right = insertIntoBST(root.right, val);
} else {
root.left = insertIntoBST(root.left, val);
}
return root;
}
}

• /**
* 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);
if (root->val < val)
root->right = insertIntoBST(root->right, val);
else
root->left = insertIntoBST(root->left, val);
return root;
}
};

• # 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: TreeNode, val: int) -> TreeNode:
def dfs(root):
if root is None:
return TreeNode(val)
if root.val < val:
root.right = dfs(root.right)
else:
root.left = dfs(root.left)
return root

return dfs(root)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func insertIntoBST(root *TreeNode, val int) *TreeNode {
if root == nil {
return &TreeNode{Val: val}
}
if root.Val < val {
root.Right = insertIntoBST(root.Right, val)
} else {
root.Left = insertIntoBST(root.Left, val)
}
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
if (!root) return new TreeNode(val);

if (val < root.val) root.left = insertIntoBST(root.left, val);
else root.right = insertIntoBST(root.right, val);

return root;
}